📖Topic Explanations

🌐 Overview
Hello students! Welcome to the fascinating world of Formation of Quadratic Equations with Given Roots! Get ready to unlock a powerful new perspective on these fundamental algebraic expressions.

You've spent a lot of time learning how to solve quadratic equations to find their roots. Now, imagine reversing that process! What if you already knew the answers (the roots) and needed to construct the original question (the quadratic equation)? This topic is precisely about that – it's like having the solution and then figuring out the problem statement that led to it.

At its core, this section teaches us how to construct a quadratic equation when its roots are provided. It might sound like a simple reversal, but it's a concept that deeply enriches your understanding of the relationship between the solutions of an equation and its algebraic form. This isn't just a mathematical trick; it's a fundamental concept that reinforces your grasp of polynomial relationships and algebraic structures.

For both your CBSE board exams and the challenging JEE Main, understanding the formation of quadratic equations from roots is absolutely crucial. It's a direct application of the relationship between roots and coefficients, a cornerstone concept that often appears in various forms in problem-solving. This knowledge allows you to:

  • Quickly form equations given specific conditions on their roots.

  • Solve problems where information about roots is given, and you need to infer the equation.

  • Develop a deeper intuition for how roots dictate the shape and form of a quadratic function.



In this section, we will delve into the elegant formulas that connect the sum of the roots and the product of the roots directly to the coefficients of the quadratic equation. You'll learn simple, yet powerful, techniques to efficiently construct any quadratic equation, whether its roots are simple integers, fractions, irrational numbers, or even complex numbers.

By mastering this concept, you won't just be able to *solve* quadratic equations; you'll gain the ability to *build* them from the ground up, giving you a far more profound and flexible command over algebraic structures. It's a step towards becoming a true architect of mathematical expressions!

So, let's embark on this exciting journey and transform our understanding of quadratic equations from merely solving them to confidently constructing them!
📚 Fundamentals
Hey everyone! Welcome back to our exciting journey through the world of Mathematics. Today, we're diving into a super interesting concept in Quadratic Equations: how to "build" a quadratic equation if you already know its solutions, which we call its roots. Think of it like being a detective! Instead of finding out who committed the crime (finding roots from an equation), we're going to solve the puzzle in reverse: if we know who the culprits are (the roots), can we reconstruct the entire story (the equation)? Absolutely!

### 1. What's a Quadratic Equation, and What are Roots? (A Quick Recap!)

Before we jump into building equations, let's quickly refresh our memory.
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable (usually 'x') is 2. Its most general form is:



$ax^2 + bx + c = 0$



Here, 'a', 'b', and 'c' are real numbers, and 'a' cannot be zero (because if 'a' was zero, it wouldn't be a quadratic equation anymore, it would be a linear one!).

The roots of a quadratic equation are simply the values of 'x' that make the equation true. They are the solutions! A quadratic equation always has exactly two roots. These roots can be real or complex, distinct or identical.

For example, if you have the equation $x^2 - 5x + 6 = 0$, you might remember factoring it as $(x-2)(x-3) = 0$. This means the roots are $x=2$ and $x=3$. If you substitute 2 or 3 back into the original equation, it will hold true!

### 2. The Basic Idea: "Reverse Engineering"

Today, we're going to do the opposite of what you usually do. Normally, you're given $ax^2 + bx + c = 0$ and asked to find 'x'. Now, we'll be given 'x' values (the roots) and asked to find $ax^2 + bx + c = 0$. It's like having the answer key and trying to figure out the original question!

Imagine you have two puzzle pieces. When you put them together correctly, they form a complete picture (the quadratic equation). These puzzle pieces are related to the roots.

Let's say the two roots of our mysterious quadratic equation are $alpha$ (alpha) and $eta$ (beta).

### 3. Method 1: Using Factors (The Intuitive Way)

This method directly uses the idea that if a number is a root, then a corresponding factor must exist.

* If $alpha$ is a root of the equation, it means when $x = alpha$, the equation becomes zero. This implies that $(x - alpha)$ must be a factor of the quadratic expression.
* Similarly, if $eta$ is a root of the equation, it means when $x = eta$, the equation becomes zero. This implies that $(x - eta)$ must also be a factor of the quadratic expression.

Since a quadratic equation has exactly two roots, it must have exactly two linear factors (ignoring constants for a moment).

So, if $alpha$ and $eta$ are the roots, the quadratic expression can be written as the product of these factors.



$(x - alpha)(x - eta) = 0$



Now, let's expand this product:
$x(x - eta) - alpha(x - eta) = 0$
$x^2 - xeta - alpha x + alphaeta = 0$


Rearranging the terms, we get:



$x^2 - (alpha + eta)x + alphaeta = 0$



Hold on, there's a small catch! What if the original equation was $2x^2 - 10x + 12 = 0$? Its roots are still 2 and 3. But if we use $(x-2)(x-3)=0$, we get $x^2 - 5x + 6 = 0$. These are clearly different equations, right?
Well, they're "equivalent" in terms of their roots. Multiplying an equation by a non-zero constant does not change its roots.
So, to be absolutely general, we should introduce a non-zero constant 'k':



$k(x - alpha)(x - eta) = 0$



or,



$k[x^2 - (alpha + eta)x + alphaeta] = 0$



where $k$ is any non-zero real number. Usually, for basic problems, we assume $k=1$ unless specified otherwise or if we need integer coefficients.

### 4. Method 2: Using Sum and Product of Roots (The Elegant Way)

This is often the preferred and more efficient method, especially for JEE problems. It connects directly to the coefficients of the general quadratic equation.

Recall the standard form:



$ax^2 + bx + c = 0$



If we divide the entire equation by 'a' (assuming $a
eq 0$), we get:



$x^2 + left(frac{b}{a}
ight)x + left(frac{c}{a}
ight) = 0$



Now, let's recall the famous relationships between roots ($alpha, eta$) and coefficients of a quadratic equation:
1. Sum of roots: $alpha + eta = -frac{b}{a}$
2. Product of roots: $alpha eta = frac{c}{a}$

Look at that! We have $b/a$ and $c/a$ in our simplified equation. Let's substitute these values:
Since $frac{b}{a} = -(alpha + eta)$, we can replace it.
And $frac{c}{a} = alphaeta$, we can replace it.

Substituting these into $x^2 + left(frac{b}{a}
ight)x + left(frac{c}{a}
ight) = 0$, we get:



$x^2 - (alpha + eta)x + alphaeta = 0$



This is the core formula you'll use most frequently!
It states that a quadratic equation with roots $alpha$ and $eta$ can be directly formed using their sum and product.

* The coefficient of 'x' is the negative of the sum of the roots.
* The constant term is the product of the roots.

Again, remember that we can multiply this entire equation by any non-zero constant $k$ to get $k[x^2 - (alpha + eta)x + alphaeta] = 0$, and the roots will remain the same.

### 5. Putting It All Together: The General Formula

So, whether you start from factors or from the sum/product relationships, you arrive at the same powerful formula.

To form a quadratic equation with given roots $alpha$ and $eta$:

1. Calculate the Sum of Roots: $S = alpha + eta$
2. Calculate the Product of Roots: $P = alpha eta$
3. Substitute into the Formula:



$x^2 - ( ext{Sum of Roots})x + ( ext{Product of Roots}) = 0$



or more compactly:



$x^2 - Sx + P = 0$



Remember, this equation is generally considered "the" quadratic equation formed. If you need integer coefficients and $S$ or $P$ are fractions, you can multiply the entire equation by a suitable constant to clear the denominators.

### 6. Let's Practice! (Examples)

Let's try a few examples to solidify our understanding. Follow along step-by-step!

Example 1: Simple Integer Roots
Form the quadratic equation whose roots are 2 and 3.


  1. Identify the roots:
    $alpha = 2$
    $eta = 3$

  2. Calculate the Sum of Roots (S):
    $S = alpha + eta = 2 + 3 = 5$

  3. Calculate the Product of Roots (P):
    $P = alpha eta = 2 imes 3 = 6$

  4. Substitute into the formula $x^2 - Sx + P = 0$:
    $x^2 - (5)x + (6) = 0$
    $x^2 - 5x + 6 = 0$


So, the quadratic equation is $x^2 - 5x + 6 = 0$. Isn't that quick and neat?

Example 2: Roots with Negative and Fractional Values
Form the quadratic equation whose roots are $-1/2$ and $4$.


  1. Identify the roots:
    $alpha = -1/2$
    $eta = 4$

  2. Calculate the Sum of Roots (S):
    $S = alpha + eta = -frac{1}{2} + 4 = -frac{1}{2} + frac{8}{2} = frac{7}{2}$

  3. Calculate the Product of Roots (P):
    $P = alpha eta = left(-frac{1}{2}
    ight) imes 4 = -2$

  4. Substitute into the formula $x^2 - Sx + P = 0$:
    $x^2 - left(frac{7}{2}
    ight)x + (-2) = 0$
    $x^2 - frac{7}{2}x - 2 = 0$

  5. (Optional but Recommended for JEE/CBSE) Clear the fractions:
    To get integer coefficients, we can multiply the entire equation by the common denominator, which is 2 in this case.
    $2 imes left(x^2 - frac{7}{2}x - 2
    ight) = 2 imes 0$
    $2x^2 - 7x - 4 = 0$


The quadratic equation is $2x^2 - 7x - 4 = 0$.

Example 3: Roots Involving Square Roots (Conjugate Pairs)
Form the quadratic equation whose roots are $1 + sqrt{2}$ and $1 - sqrt{2}$.


  1. Identify the roots:
    $alpha = 1 + sqrt{2}$
    $eta = 1 - sqrt{2}$
    Notice these are "conjugate pairs". This often happens when coefficients are rational.

  2. Calculate the Sum of Roots (S):
    $S = alpha + eta = (1 + sqrt{2}) + (1 - sqrt{2})$
    $S = 1 + sqrt{2} + 1 - sqrt{2}$
    $S = 2$ (The $sqrt{2}$ terms cancel out, simplifying things!)

  3. Calculate the Product of Roots (P):
    $P = alpha eta = (1 + sqrt{2})(1 - sqrt{2})$
    This is in the form $(a+b)(a-b) = a^2 - b^2$.
    $P = (1)^2 - (sqrt{2})^2$
    $P = 1 - 2$
    $P = -1$

  4. Substitute into the formula $x^2 - Sx + P = 0$:
    $x^2 - (2)x + (-1) = 0$
    $x^2 - 2x - 1 = 0$


The quadratic equation is $x^2 - 2x - 1 = 0$. See how even with tricky-looking roots, the sum and product method makes it quite straightforward?

### 7. Why Bother Learning This? (Relevance)

You might be thinking, "Why do I need to know how to form an equation from its roots?" This skill is super useful in many areas:

* Problem Solving: Many JEE problems provide information about the roots (their sum, product, or some relationship between them) and ask you to find the equation or relate its coefficients. Knowing this direct formation method is crucial.
* Connecting Concepts: It reinforces the deep connection between the roots of an equation and its coefficients.
* Advanced Topics: This fundamental understanding extends to higher-degree polynomials (cubic, quartic equations) where similar relationships exist between roots and coefficients.
* Checking Answers: If you've found roots for an equation, you can quickly form an equation using those roots and see if it matches the original equation, giving you confidence in your answer.

So, mastering this "reverse engineering" technique is not just about memorizing a formula; it's about understanding the underlying structure of quadratic equations. Keep practicing, and you'll find it becomes second nature!
🔬 Deep Dive
Hello, future IITians! Welcome to this "Deep Dive" session on one of the fundamental yet crucial topics in Quadratic Equations: Formation of Quadratic Equations with Given Roots. While it might seem straightforward, understanding its nuances and applications, especially for JEE, is paramount. We'll start from the very basics and build up to advanced transformation techniques that are frequently tested in competitive exams.

### 1. The Genesis: Understanding Roots and Factors

Before we form an equation, let's quickly recall what roots are. For a quadratic equation $ax^2 + bx + c = 0$ (where $a
eq 0$), the roots (or zeroes) are the values of $x$ that satisfy the equation. Geometrically, they are the x-intercepts of the parabola $y = ax^2 + bx + c$.

If $alpha$ and $eta$ are the roots of a quadratic equation, it means that when $x = alpha$ or $x = eta$, the expression $ax^2+bx+c$ becomes zero. This implies that $(x - alpha)$ and $(x - eta)$ must be factors of the quadratic expression.

Think about it this way: if a number 5 is a root of an equation, then $(x-5)$ must be a factor. If 3 is another root, then $(x-3)$ is also a factor. Therefore, the quadratic expression must be proportional to the product of these factors.

So, a quadratic equation whose roots are $alpha$ and $eta$ can be written in the form:
$k(x - alpha)(x - eta) = 0$
where $k$ is any non-zero real constant.

Let's expand this:
$k(x - alpha)(x - eta) = 0$
$k[x(x - eta) - alpha(x - eta)] = 0$
$k[x^2 - eta x - alpha x + alpha eta] = 0$
$k[x^2 - (alpha + eta)x + alpha eta] = 0$

This leads us to the most fundamental form for forming a quadratic equation:
$x^2 - ( ext{Sum of Roots})x + ( ext{Product of Roots}) = 0$
Or, more generally, including the constant $k$:
$k[x^2 - ( ext{Sum of Roots})x + ( ext{Product of Roots})] = 0$

Important Note for JEE: The inclusion of the constant 'k' is crucial. If a question asks for "a quadratic equation," then the $k=1$ form is acceptable. However, if it asks for "the quadratic equation" satisfying certain conditions (e.g., specific leading coefficient, or passing through a certain point), then 'k' becomes significant and needs to be determined. Remember, multiplying an equation by a non-zero constant does not change its roots. So, there are infinitely many quadratic equations sharing the same roots!

### 2. Connecting to Vieta's Formulas

Recall Vieta's formulas for a standard quadratic equation $ax^2 + bx + c = 0$:
Sum of roots: $alpha + eta = -b/a$
Product of roots: $alpha eta = c/a$

We can rewrite $ax^2 + bx + c = 0$ by dividing by 'a' (assuming $a
eq 0$):
$x^2 + (b/a)x + (c/a) = 0$
Substituting Vieta's formulas:
$x^2 - (- ext{Sum of Roots})x + ( ext{Product of Roots}) = 0$
$x^2 - (alpha + eta)x + alpha eta = 0$

This confirms our derived formula. This is the monic quadratic equation (where the coefficient of $x^2$ is 1) with roots $alpha$ and $eta$.

---

### 3. Step-by-Step Examples

Let's put this into practice with a few examples covering different types of roots.

Example 1: Real and Distinct Roots
Form a quadratic equation whose roots are 2 and -5.

Solution:
Let $alpha = 2$ and $eta = -5$.
1. Calculate the Sum of Roots:
$alpha + eta = 2 + (-5) = -3$
2. Calculate the Product of Roots:
$alpha eta = (2)(-5) = -10$
3. Form the Equation:
Using the formula $x^2 - ( ext{Sum})x + ( ext{Product}) = 0$:
$x^2 - (-3)x + (-10) = 0$
$x^2 + 3x - 10 = 0$

We can verify this by factoring: $(x+5)(x-2)=0$, which yields roots $x=-5$ and $x=2$.

Example 2: Irrational Conjugate Roots
Form a quadratic equation whose roots are $3 + sqrt{2}$ and $3 - sqrt{2}$.

Solution:
Let $alpha = 3 + sqrt{2}$ and $eta = 3 - sqrt{2}$.
1. Calculate the Sum of Roots:
$alpha + eta = (3 + sqrt{2}) + (3 - sqrt{2}) = 3 + 3 + sqrt{2} - sqrt{2} = 6$
2. Calculate the Product of Roots:
$alpha eta = (3 + sqrt{2})(3 - sqrt{2})$
This is in the form $(a+b)(a-b) = a^2 - b^2$:
$alpha eta = 3^2 - (sqrt{2})^2 = 9 - 2 = 7$
3. Form the Equation:
$x^2 - ( ext{Sum})x + ( ext{Product}) = 0$
$x^2 - 6x + 7 = 0$

JEE Insight (Conjugate Root Theorem): If a quadratic equation with rational coefficients has one irrational root of the form $a+sqrt{b}$ (where $b$ is not a perfect square), then its conjugate $a-sqrt{b}$ must also be a root. This is why when you're given just one irrational root and told coefficients are rational, you automatically know the other root.

Example 3: Complex Conjugate Roots
Form a quadratic equation whose roots are $1 + 2i$ and $1 - 2i$.

Solution:
Let $alpha = 1 + 2i$ and $eta = 1 - 2i$.
1. Calculate the Sum of Roots:
$alpha + eta = (1 + 2i) + (1 - 2i) = 1 + 1 + 2i - 2i = 2$
2. Calculate the Product of Roots:
$alpha eta = (1 + 2i)(1 - 2i)$
Again, using $(a+b)(a-b) = a^2 - b^2$:
$alpha eta = 1^2 - (2i)^2 = 1 - (4i^2)$
Since $i^2 = -1$:
$alpha eta = 1 - (4 imes -1) = 1 - (-4) = 1 + 4 = 5$
3. Form the Equation:
$x^2 - ( ext{Sum})x + ( ext{Product}) = 0$
$x^2 - 2x + 5 = 0$

JEE Insight (Conjugate Root Theorem for Complex Numbers): If a quadratic equation with real coefficients has one complex root of the form $a+bi$ (where $b
eq 0$), then its conjugate $a-bi$ must also be a root. This is a very common scenario in JEE problems.

---

### 4. Advanced Applications: Root Transformations (JEE Focus!)

This is where the concept gets interesting for JEE. Often, you're given an original quadratic equation and asked to form a new equation whose roots are some function of the roots of the original equation.

Let $alpha, eta$ be the roots of the equation $ax^2 + bx + c = 0$. We want to find a new quadratic equation whose roots are $alpha'$ and $eta'$, where $alpha'$ and $eta'$ are related to $alpha$ and $eta$.

There are two primary methods:

#### Method 1: Using Sum and Product of New Roots

This method involves directly calculating the sum ($alpha' + eta'$) and product ($alpha' eta'$) of the new roots and then using the formula $x^2 - (alpha' + eta')x + alpha' eta' = 0$. This is generally more straightforward for symmetric transformations (where $alpha'$ and $eta'$ are the same function of $alpha$ and $eta$ respectively, e.g., $alpha^2, eta^2$).

Example 4: Roots are Squares of Original Roots
If $alpha, eta$ are the roots of $x^2 - 5x + 6 = 0$, form a quadratic equation whose roots are $alpha^2, eta^2$.

Solution:
For $x^2 - 5x + 6 = 0$:
Sum of roots: $alpha + eta = -(-5)/1 = 5$
Product of roots: $alpha eta = 6/1 = 6$

New roots are $alpha' = alpha^2$ and $eta' = eta^2$.
1. New Sum of Roots:
$alpha' + eta' = alpha^2 + eta^2$
We know that $alpha^2 + eta^2 = (alpha + eta)^2 - 2alphaeta$.
So, $alpha^2 + eta^2 = (5)^2 - 2(6) = 25 - 12 = 13$.
2. New Product of Roots:
$alpha' eta' = alpha^2 eta^2 = (alpha eta)^2 = (6)^2 = 36$.
3. Form the New Equation:
$x^2 - ( ext{New Sum})x + ( ext{New Product}) = 0$
$x^2 - 13x + 36 = 0$

#### Method 2: Using Transformation/Substitution (Very Powerful for JEE)

This method is particularly useful when the transformation is complex or involves non-symmetric functions.
Let $y$ be a new root. Express $y$ in terms of $x$ (an old root). Then, express $x$ in terms of $y$. Substitute this expression for $x$ back into the original quadratic equation. The resulting equation in $y$ will be the new quadratic equation.

Example 5: Roots are Reciprocals of Original Roots
If $alpha, eta$ are the roots of $ax^2 + bx + c = 0$, form a quadratic equation whose roots are $1/alpha, 1/eta$.

Solution:
Let $y$ be a new root. So, $y = 1/x$.
From this, we can express $x$ in terms of $y$: $x = 1/y$.

Now, substitute $x = 1/y$ into the original equation $ax^2 + bx + c = 0$:
$a(1/y)^2 + b(1/y) + c = 0$
$a/y^2 + b/y + c = 0$

To eliminate denominators, multiply the entire equation by $y^2$ (assuming $y
eq 0$, which means $alpha, eta
eq 0$; if $c=0$, then one root is zero, and its reciprocal is undefined, so we assume $c
eq 0$):
$a + by + cy^2 = 0$
Rearranging into standard quadratic form:
$cy^2 + by + a = 0$
Or, replacing $y$ with $x$ as the variable: $cx^2 + bx + a = 0$

Notice how the coefficients just reversed! This is a classic JEE result.

Let's try Example 4 again using this substitution method for practice.

Example 6: Roots are Squares of Original Roots (using substitution)
If $alpha, eta$ are the roots of $x^2 - 5x + 6 = 0$, form a quadratic equation whose roots are $alpha^2, eta^2$.

Solution:
Let $y$ be a new root. So, $y = x^2$.
We need to express $x$ in terms of $y$. Here, $x = pm sqrt{y}$. This implies there might be issues with uniqueness if not handled carefully, or that a new equation in $y$ will naturally square out.

Let's rearrange the original equation to isolate terms involving $x$:
$x^2 = 5x - 6$
Since we want the new root to be $y=x^2$, we can directly substitute $y$ for $x^2$:
$y = 5x - 6$
Now we need to eliminate $x$. From this equation, $5x = y+6$, so $x = (y+6)/5$.
Substitute this $x$ back into the *original* equation $x^2 - 5x + 6 = 0$:
$((y+6)/5)^2 - 5((y+6)/5) + 6 = 0$
$(y+6)^2/25 - (y+6) + 6 = 0$
$(y^2 + 12y + 36)/25 - (y+6) + 6 = 0$
Multiply by 25 to clear the denominator:
$y^2 + 12y + 36 - 25(y+6) + 25(6) = 0$
$y^2 + 12y + 36 - 25y - 150 + 150 = 0$
$y^2 + (12-25)y + (36 - 150 + 150) = 0$
$y^2 - 13y + 36 = 0$
Which, replacing $y$ with $x$, is $x^2 - 13x + 36 = 0$. This matches the result from Method 1. This method is generally more robust for complex transformations.

Example 7: Roots shifted by a constant
If $alpha, eta$ are the roots of $x^2 - 7x + 10 = 0$, form a quadratic equation whose roots are $alpha+3, eta+3$.

Solution (Method 1: Sum & Product):
Original: $alpha+eta=7$, $alphaeta=10$.
New roots: $alpha' = alpha+3$, $eta' = eta+3$.
New Sum: $alpha'+eta' = (alpha+3) + (eta+3) = (alpha+eta) + 6 = 7+6=13$.
New Product: $alpha'eta' = (alpha+3)(eta+3) = alphaeta + 3alpha + 3eta + 9 = alphaeta + 3(alpha+eta) + 9$.
$= 10 + 3(7) + 9 = 10 + 21 + 9 = 40$.
New equation: $x^2 - 13x + 40 = 0$

Solution (Method 2: Substitution):
Let $y$ be a new root. $y = x+3$.
Then $x = y-3$.
Substitute $x=y-3$ into $x^2 - 7x + 10 = 0$:
$(y-3)^2 - 7(y-3) + 10 = 0$
$(y^2 - 6y + 9) - (7y - 21) + 10 = 0$
$y^2 - 6y + 9 - 7y + 21 + 10 = 0$
$y^2 - 13y + 40 = 0$
Replacing $y$ with $x$: $x^2 - 13x + 40 = 0$.

This demonstrates the elegance and power of the substitution method for general root transformations.

---

### 5. CBSE vs. JEE Focus































Aspect CBSE (Class 10/11) JEE Main & Advanced
Basic Formation Direct application of $x^2 - ( ext{Sum})x + ( ext{Product}) = 0$. Roots are usually simple integers or surds (irrational numbers). Complex roots are introduced later (Class 11/12). Foundationally same, but roots can be complex, involving $i$, or more intricate surds. The constant 'k' might be important if an additional condition is given (e.g., passes through a point, specific leading coefficient).
Root Transformations Typically involves finding equations with roots like $1/alpha, 1/eta$ or $alpha+k, eta+k$ by directly calculating the new sum and product. Problems are less intricate. Extensive use of root transformations. Students are expected to be proficient with both the direct sum/product method and the substitution method. Transformations can involve higher powers ($alpha^3, eta^3$), rational functions ($(alpha+1)/(alpha-1)$), or combinations. Questions can be layered, requiring manipulation of algebraic identities for symmetric polynomials.
Conjugate Root Theorem Introduction to the concept for irrational and complex roots when coefficients are real/rational. Implicitly assumed. Used frequently to deduce the "missing" root if one is given, especially in polynomial equations of higher degrees, making it crucial for forming the quadratic factor.
Problem Complexity Straightforward, direct application of formulas. Focus on understanding the relationship between roots and coefficients. Multi-step problems, often combining formation of equations with other concepts like properties of roots, common roots, graphs, or calculus. Requires strong algebraic manipulation skills and conceptual clarity.


---

### Conclusion

Forming quadratic equations from given roots is a foundational skill. For CBSE, mastery of the basic formula and simple transformations is sufficient. However, for JEE, you must delve deeper into root transformations, mastering both the direct sum/product method and the powerful substitution technique. Understanding the role of the constant 'k' and the conjugate root theorems are also vital. Practice with diverse problem types will solidify your understanding and prepare you for advanced challenges. Keep practicing, and you'll master this topic!
🎯 Shortcuts

Mastering the formation of quadratic equations from given roots is crucial for both JEE Main and CBSE board exams. Here are some effective mnemonics and shortcuts to help you remember the key formula and speed up your calculations.



1. The "S.P. Formula" Mnemonic


The most fundamental shortcut for forming a quadratic equation when its roots (let's say $alpha$ and $eta$) are given is to remember its general form:


$x^2 - ( ext{Sum of roots})x + ( ext{Product of roots}) = 0$


To remember this, think of the "S.P. Formula":



  • S stands for Sum of roots ($alpha + eta$)

  • P stands for Product of roots ($alpha eta$)


So, the formula becomes: $x^2 - Sx + P = 0$


This is a very direct and widely used mnemonic that simplifies recalling the structure of the equation.



2. Sign Pattern Mnemonic: "SAP" or "Plus-Minus-Plus"


Students often make mistakes with the signs in the formula. Remember this simple pattern:



  • S.A.P. Mnemonic: "Subtract the Sum, Add the Product."

    • The $x$ term (associated with the Sum of roots) is always subtracted.

    • The constant term (associated with the Product of roots) is always added.



  • Plus-Minus-Plus Pattern: Observe the signs for the terms in the standard form $Ax^2 + Bx + C = 0$ when $A=1$:

    • $x^2$ term: Implicitly positive (+)

    • $x$ term: Negative (-) for the sum

    • Constant term: Positive (+) for the product


    So, it's a + - + sign pattern for $x^2$, $x$, and the constant term, respectively.





3. Shortcut for Complex Conjugate Roots (JEE Tip)


If the roots are complex conjugates, say $alpha = a + ib$ and $eta = a - ib$, calculating their sum and product quickly is a significant time-saver in JEE:



  • Sum of Roots (S):

    $S = (a + ib) + (a - ib) = 2a$


    Shortcut: Just double the real part.



  • Product of Roots (P):

    $P = (a + ib)(a - ib) = a^2 - (ib)^2 = a^2 + b^2$


    Shortcut: Sum of the squares of the real and imaginary parts (ignoring $i$).




Example Quick Application: If roots are $2 + 3i$ and $2 - 3i$:



  • $S = 2 imes 2 = 4$

  • $P = 2^2 + 3^2 = 4 + 9 = 13$


The equation is $x^2 - 4x + 13 = 0$. This avoids direct addition and multiplication of complex numbers, saving time.



4. General Speed Tip (CBSE & JEE)


Whenever you are given roots, your immediate first step should be to calculate their Sum and Product. Make this a habit. Once you have these two values, simply plug them into the $x^2 - Sx + P = 0$ formula. This systematic approach ensures accuracy and speed.


Stay sharp and practice these shortcuts to make forming quadratic equations effortless!

💡 Quick Tips

Quick Tips for Forming Quadratic Equations


Forming a quadratic equation when its roots are given is a fundamental skill, directly applying Vieta's formulas. Mastering this concept ensures quick solutions to related problems in both board exams and JEE Main.





  • The Golden Rule (Fundamental Formula):

    The most crucial tip is to remember the standard form for a quadratic equation given its roots α and β:


    x2 - (Sum of Roots)x + (Product of Roots) = 0


    Or, more formally: x2 - (α + β)x + (αβ) = 0.


    Always start here. Calculate the sum (α + β) and the product (αβ) of the given roots first, then substitute them into this formula.




  • Handling the Leading Coefficient 'a':

    The formula x2 - (S)x + (P) = 0 assumes the leading coefficient (coefficient of x2) is 1. If the question specifies that the quadratic equation should have a specific leading coefficient 'a' (e.g., "form a quadratic equation with roots 2 and 3, and leading coefficient 5"), then the equation becomes:


    a[x2 - (Sum of Roots)x + (Product of Roots)] = 0


    Or, a[x2 - (α + β)x + (αβ)] = 0. Remember to multiply the entire expression by 'a'.




  • Conjugate Root Pairs (JEE Main Focus):

    This is extremely important for JEE Main and slightly more complex problems:




    • Irrational Roots: If one root is given as p + &sqrt;q (where p and q are rational, and &sqrt;q is irrational), then its conjugate, p - &sqrt;q, must be the other root, provided the coefficients of the quadratic equation are rational.


    • Complex Roots: If one root is given as p + iq (where p and q are real, and q ≠ 0), then its complex conjugate, p - iq, must be the other root, provided the coefficients of the quadratic equation are real.


    This property allows you to determine the second root immediately if only one irrational or complex root is provided in such contexts.




  • Simplification and Integer Coefficients:

    After forming the equation, always simplify it to its lowest integer coefficients. For example, if you get `2x^2 - 4x + 6 = 0`, simplify it to `x^2 - 2x + 3 = 0` by dividing by 2 (unless a specific leading coefficient is asked for). This is standard practice in both CBSE and JEE.




  • Verification (Quick Check):

    If time permits, quickly substitute one of the given roots back into your derived quadratic equation. If it satisfies the equation (makes it zero), your equation is likely correct. This helps catch calculation errors in sum or product.




  • CBSE vs. JEE Approach:

    For CBSE Board Exams, roots are typically simple integers or rational numbers. The direct application of the formula x2 - Sx + P = 0 is usually sufficient.


    For JEE Main, expect roots to be irrational, complex, or involving parameters. The conjugate root pair property becomes crucial here for efficiency and accuracy. Questions might also involve relating the formed equation to another quadratic equation or conditions.




Stay focused, practice carefully with different types of roots, and these tips will help you quickly form quadratic equations!

🧠 Intuitive Understanding

Intuitive Understanding: Forming Quadratic Equations from Roots



Understanding how to construct a quadratic equation when its roots are known is a fundamental concept in algebra. It relies on a very simple, yet powerful, idea: If a number is a root of an equation, then substituting that number into the equation makes it true (i.e., makes the equation evaluate to zero).

Let's break down the intuition:

1. Roots and Factors – The Core Connection:
* Consider a simple linear equation: `x - a = 0`. What is its root? Clearly, `x = a`.
* This means if `x = a` is a root of *any* polynomial equation, then `(x - a)` must be a factor of that polynomial. Why? Because when you substitute `x = a`, the factor `(a - a)` becomes `0`, and anything multiplied by `0` is `0`, thus satisfying the equation.
* Conversely, if `(x - a)` is a factor, then setting it to zero gives `x - a = 0`, so `x = a` is a root.

2. Extending to Two Roots for a Quadratic Equation:
* A quadratic equation has exactly two roots (which may be real or complex, and distinct or identical). Let these roots be $alpha$ (alpha) and $eta$ (beta).
* Based on our understanding above, if $alpha$ is a root, then `(x - $alpha$)` must be a factor of the quadratic polynomial.
* Similarly, if $eta$ is a root, then `(x - $eta$)` must also be a factor of the quadratic polynomial.
* For a polynomial to have both `(x - $alpha$)` and `(x - $eta$)` as factors, the simplest way is for the polynomial itself to be the product of these factors.

3. Constructing the Equation:
* Therefore, the most basic quadratic equation with roots $alpha$ and $eta$ can be written as:


`$(x - alpha)(x - eta) = 0$`


* This equation intuitively makes sense: if you substitute `x = $alpha$`, the first factor becomes `0`, making the whole equation `0`. If you substitute `x = $eta$`, the second factor becomes `0`, again making the whole equation `0`.

4. The Role of a Constant Multiplier 'k':
* While `$(x - alpha)(x - eta) = 0$` gives *a* quadratic equation, it's not the *only* one. For example, `(x - 1)(x - 2) = 0` and `2(x - 1)(x - 2) = 0` both have roots `x=1` and `x=2`.
* This implies that any non-zero constant `k` can multiply the factored form without changing the roots.
* So, the general form of a quadratic equation with roots $alpha$ and $eta$ is:


`$k(x - alpha)(x - eta) = 0$`, where `$k
eq 0$`.

5. Connecting to Sum and Product of Roots (JEE & CBSE Relevance):
* Expanding the general form `$(x - alpha)(x - eta) = 0$` leads to:
`$x^2 - eta x - alpha x + alpha eta = 0$`
`$x^2 - (alpha + eta)x + alpha eta = 0$`
* This is the standard form `$x^2 - ( ext{Sum of Roots})x + ( ext{Product of Roots}) = 0$`. This intuitive derivation shows that the "sum and product of roots" formula is not just a memorized rule, but a direct consequence of the fundamental factor-root relationship. Both CBSE and JEE extensively test this concept. For JEE, this understanding is crucial for complex problems involving root transformations or relationships between coefficients.

Example:
If the roots of a quadratic equation are 3 and -2:
1. Intuitive Step: Since 3 is a root, `(x - 3)` is a factor. Since -2 is a root, `(x - (-2))` which is `(x + 2)` is a factor.
2. Form the Equation: `$(x - 3)(x + 2) = 0$`
3. Expand: `$x^2 + 2x - 3x - 6 = 0 implies x^2 - x - 6 = 0$`
(We can also have `$k(x^2 - x - 6) = 0$` for any `$k
eq 0$`).

This intuitive understanding ensures you can reconstruct a quadratic equation from its roots even if you momentarily forget the specific sum/product formula, by returning to the core definition of a root.
🌍 Real World Applications

Real-World Applications of Forming Quadratic Equations from Roots


While often perceived as an abstract mathematical concept, the ability to form a quadratic equation when its roots (or solutions) are known has significant applications across various scientific, engineering, and economic fields. Understanding these applications helps in appreciating the practical utility of quadratic equations beyond theoretical problem-solving.



Core Concept Reminder:


Recall that if α and β are the roots of a quadratic equation, the equation can be expressed as:

x² - (α + β)x + αβ = 0

or equivalently, as k(x - α)(x - β) = 0, where 'k' is a non-zero constant.



Key Application Areas:




  • Physics and Engineering (Projectile Motion):

    In projectile motion, the path of an object (like a ball thrown in the air) often follows a parabolic trajectory, which can be described by a quadratic equation. If you know the exact points where the projectile starts and lands (e.g., initial height = 0 at time t=0, and landing height = 0 at time t=T), these serve as the roots of the height function h(t). Knowing these roots allows engineers or physicists to form the specific quadratic equation describing the object's height over time, even if other parameters (like initial velocity) are not directly given. This is crucial for trajectory prediction, design of sports equipment, or ballistic calculations.




  • Architecture and Design (Parabolic Structures):

    Parabolic shapes are frequently used in architecture (e.g., bridge arches, domes) and optics (e.g., satellite dishes, car headlights). If the start and end points of a parabolic arch are fixed (e.g., where it meets the ground), these points can be considered as the roots of the quadratic equation describing the arch's cross-section. Architects can then use these roots to define the equation, ensuring the structure spans the desired distance and has the correct parabolic form. The constant 'k' in k(x-α)(x-β)=0 would then determine the height or 'openness' of the arch.




  • Economics and Business (Profit/Cost Functions):

    In business, quadratic functions are often used to model profit or cost. For instance, a profit function P(x) = -ax² + bx + c might show that profit is zero at certain production levels (break-even points). If a business aims for specific break-even points, say at x units and y units, these values would be the roots of the profit function when P(x)=0. The company can then *construct* a quadratic profit function that inherently has these desired break-even points, and subsequently optimize production to maximize profit within these boundaries.





JEE/CBSE Relevance:


While direct "real-world application" problems where you *form* an equation from roots for a physical scenario are less common in JEE and CBSE exams, understanding these applications:



  • Deepens Conceptual Understanding: It reinforces why quadratic equations are important and how their properties (like roots) translate into practical contexts.

  • Enhances Problem-Solving Skills: It encourages a more holistic view of mathematical problems, connecting abstract algebra to tangible situations.


The core skill of forming quadratic equations from given roots is fundamental and directly tested in both board exams and competitive entrance tests.


Keep connecting the dots between theory and application – it makes learning more engaging and meaningful!

🔄 Common Analogies

Common Analogies for Forming Quadratic Equations



Understanding how to form a quadratic equation when its roots are given can be made simpler by drawing parallels with more familiar concepts. These analogies help solidify the underlying logic behind the formula $x^2 - ( ext{Sum of roots})x + ( ext{Product of roots}) = 0$.

Here are some common analogies:





JEE Tip: While analogies aid conceptual understanding, for JEE, mastering the direct application of the sum and product of roots formula is key for speed and accuracy.







  1. The "Reverse Engineering" of Factors:


    • The Forward Process (Solving): Imagine you have a locked box (the quadratic equation) and you need to find what's inside (the roots). You factor the equation, breaking it down into two simpler parts $(x-alpha)(x-eta)=0$, and then you find the values of $x$ (the roots $alpha$ and $eta$) that make each part zero. This is like dismantling a machine to see how it works.


    • The Reverse Process (Forming): Now, if you are given the contents (the roots $alpha$ and $eta$), and you need to rebuild the original box (the quadratic equation), you simply work backwards.

      • If $alpha$ is a root, then $(x-alpha)$ must be a factor.

      • If $eta$ is a root, then $(x-eta)$ must be a factor.


      To reconstruct the original quadratic equation, you just multiply these factors:

      (x - $alpha$)(x - $eta$) = 0


      Expanding this gives you $x^2 - (alpha+eta)x + alphaeta = 0$, which is the standard form involving the sum and product of roots. This is like assembling a machine when you know its components.




  2. The "Chef's Recipe" for a Quadratic Dish:


    • Ingredients (Roots): Think of the given roots, say $alpha$ and $eta$, as your primary ingredients for a special mathematical dish (the quadratic equation). You have these two core elements.


    • The Recipe (Formula): To create the dish, you don't just throw the ingredients together randomly. You need a specific recipe. The formula $x^2 - ( ext{Sum of roots})x + ( ext{Product of roots}) = 0$ acts as that precise recipe.


    • Combining Ingredients (Sum & Product): The recipe instructs you to first combine your main ingredients in two specific ways:

      • Calculate their Sum ($alpha + eta$). This is one specific "preparation" of your ingredients.

      • Calculate their Product ($alpha cdot eta$). This is another specific "preparation."




    • Assembling the Dish: Now, you follow the recipe card: "Take $x^2$, subtract the 'sum-prepared' ingredients multiplied by $x$, then add the 'product-prepared' ingredients, and set the whole thing to zero."

      $x^2 - ( ext{Sum})x + ( ext{Product}) = 0$


      The exact structure, including the crucial minus sign before the sum term, is like a precise cooking step that ensures the dish turns out correctly.



📋 Prerequisites

Before diving into the formation of quadratic equations from given roots, it is essential to have a strong foundation in some core concepts related to quadratic expressions and equations. Mastering these prerequisites will ensure a smoother understanding of the target topic and its applications in problem-solving.



Here are the key prerequisites:





  • Standard Form of a Quadratic Equation:

    • You must be thoroughly familiar with the general form of a quadratic equation: ax² + bx + c = 0, where 'a', 'b', and 'c' are real numbers, and a ≠ 0.

    • Understand the roles of the coefficients 'a', 'b', and 'c' and how they define the equation.




  • Understanding Roots of a Quadratic Equation:

    • Know that the roots (or solutions) of a quadratic equation are the values of the variable (usually 'x') that satisfy the equation, i.e., make the equation true.

    • A quadratic equation typically has two roots, which can be real, distinct, real and equal, or complex conjugates.

    • Understand that if $alpha$ and $eta$ are the roots of ax² + bx + c = 0, then (x - $alpha$) and (x - $eta$) are factors of the quadratic expression ax² + bx + c.




  • Relationship Between Roots and Coefficients (Vieta's Formulas for Quadratics):

    • This is arguably the most crucial prerequisite for this topic. You must know and be able to apply Vieta's formulas for quadratic equations.

    • For a quadratic equation ax² + bx + c = 0 with roots $alpha$ and $eta$:

      • Sum of Roots: $alpha$ + $eta$ = -b/a

      • Product of Roots: $alpha$$eta$ = c/a



    • JEE Focus: Mastery of these relationships is fundamental for solving a wide array of problems in quadratic equations, beyond just forming them. It's often used in conjunction with other algebraic identities.




  • Basic Algebraic Simplification and Expansion:

    • Proficiency in expanding algebraic expressions, particularly products like (x - $alpha$)(x - $eta$).

    • Ability to perform basic addition, subtraction, multiplication, and division of algebraic terms and expressions.

    • Knowledge of common algebraic identities, such as (a+b)², (a-b)², etc., can be helpful for simplifying expressions involving roots.





By ensuring a solid understanding of these concepts, you will be well-prepared to grasp the methods for forming quadratic equations when their roots are provided.

🔗 Related Topics

Related Topics: Formation of Quadratic Equations with Given Roots



Understanding how to form a quadratic equation from its roots is deeply interconnected with several fundamental concepts in algebra. Mastering these related topics will not only solidify your grasp on quadratic equations but also prepare you for more advanced problems in polynomial theory.



  • Vieta's Formulas (Relations between Roots and Coefficients):

    This is the most direct and crucial related topic. Vieta's formulas provide the foundation for forming quadratic equations. They state that for a quadratic equation (ax^2 + bx + c = 0) with roots (alpha) and (eta):



    • Sum of roots: (alpha + eta = -frac{b}{a})

    • Product of roots: (alpha eta = frac{c}{a})


    Conversely, if you know the sum (S) and product (P) of the roots, the quadratic equation can be written as (x^2 - Sx + P = 0). This is the core principle behind forming equations from given roots.


    JEE Main Relevance: Vieta's formulas are indispensable for solving problems involving relationships between roots and coefficients, especially when dealing with symmetric expressions of roots or conditions on the roots.




  • Nature of Roots:

    The discriminant ((Delta = b^2 - 4ac)) determines whether the roots are real or complex, rational or irrational. When forming an equation, if roots are given, their nature influences how you process them, particularly with non-real or irrational roots. For example, if one root is (a + sqrt{b}) (where (sqrt{b}) is irrational) or (a + ib), its conjugate ((a - sqrt{b}) or (a - ib) respectively) must also be a root if the coefficients of the quadratic equation are rational or real, respectively.


    CBSE vs JEE: Both curricula cover the nature of roots. JEE often integrates this concept into more complex problem-solving scenarios, especially with complex roots.




  • Symmetric Functions of Roots:

    Expressions that remain unchanged when the roots are interchanged (e.g., (alpha^2 + eta^2), (frac{1}{alpha} + frac{1}{eta})) are called symmetric functions. These can always be expressed in terms of the elementary symmetric functions, which are (alpha + eta) and (alpha eta). This skill is crucial for problems where you need to find an equation whose roots are some transformation of the roots of a given equation.


    For example: (alpha^2 + eta^2 = (alpha + eta)^2 - 2alphaeta)




  • Transformations of Equations / Roots:

    This is an advanced application where you are given a quadratic equation (f(x) = 0) with roots (alpha, eta), and you need to find a new quadratic equation whose roots are, for instance, ((alpha+k), (eta+k)) or (frac{1}{alpha}, frac{1}{eta}) or (alpha^2, eta^2). This often involves using Vieta's formulas to find the sum and product of the new roots in terms of the sum and product of the original roots, or by a substitution method.


    For example, if roots are (2alpha, 2eta), the new sum is (2(alpha+eta)) and new product is (4alphaeta).


    JEE Main Relevance: Problems on transformations of roots are very common in JEE Main, testing your ability to manipulate expressions involving roots and apply Vieta's formulas effectively.




  • Polynomial Equations of Higher Degree:

    The concept of forming an equation from roots is generalized to polynomials of higher degrees (cubic, quartic, etc.). Vieta's formulas also extend to these polynomials, relating the sums and products of roots taken one at a time, two at a time, and so on, to the coefficients of the polynomial. Understanding the quadratic case forms a strong base for these generalizations.




By studying these related topics, you'll develop a holistic understanding of quadratic equations, moving beyond just mechanical calculations to a deeper conceptual and problem-solving level.

⚠️ Common Exam Traps

When forming quadratic equations from given roots, students often fall into specific traps during exams. Being aware of these common pitfalls can significantly improve accuracy and prevent loss of marks, especially in competitive exams like JEE Main.



Common Exam Traps for Forming Quadratic Equations





  • Sign Errors in the Sum of Roots:

    The standard form of a quadratic equation with roots $alpha$ and $eta$ is $x^2 - (alpha + eta)x + alphaeta = 0$. A very common mistake is to write $x^2 + (alpha + eta)x + alphaeta = 0$. Always remember the negative sign before the sum of roots term.


    JEE Tip: This fundamental error can be costly. Double-check the sign, especially when roots are negative or complex.




  • Forgetting Conjugate Roots (Crucial for JEE):

    If the coefficients of a quadratic equation are real, and one root is irrational (e.g., $a+sqrt{b}$) or complex (e.g., $c+id$), then its conjugate must also be a root.


    • If one root is $2+sqrt{3}$, and coefficients are real, the other root must be $2-sqrt{3}$.

    • If one root is $1+2i$, and coefficients are real, the other root must be $1-2i$.


    Many problems will give only one such root and imply the real coefficient condition. Failing to assume the conjugate root will lead to incorrect sum and product of roots.


    JEE Specific: This is a high-frequency trap in JEE. Always check if the problem states or implies "real coefficients" when given a single irrational or complex root.




  • Ignoring the Leading Coefficient:

    The formula $x^2 - (alpha + eta)x + alphaeta = 0$ assumes the leading coefficient (coefficient of $x^2$) is 1. If the question specifies that the quadratic equation has a leading coefficient 'a' (where $a
    eq 1$), or asks for "a quadratic equation" and later uses it in a context where $a
    eq 1$, you must multiply the entire equation by 'a'. The general form is $a[x^2 - (alpha + eta)x + alphaeta] = 0$.


    Example: If roots are 2 and 3, and the leading coefficient is 5, the equation is $5(x^2 - (2+3)x + (2)(3)) = 0 Rightarrow 5(x^2 - 5x + 6) = 0 Rightarrow 5x^2 - 25x + 30 = 0$. Not just $x^2 - 5x + 6 = 0$.




  • Careless Calculation of Sum and Product:

    When roots involve fractions, radicals, or complex numbers, errors often occur during the calculation of their sum ($alpha+eta$) and product ($alphaeta$).



    • Fractions: Ensure common denominators are correctly applied.

    • Radicals: Remember properties like $(sqrt{a}+sqrt{b})(sqrt{a}-sqrt{b}) = a-b$.

    • Complex Numbers: Utilize properties like $(a+ib)(a-ib) = a^2+b^2$.


    A small arithmetic error here will lead to an entirely incorrect quadratic equation.





By being mindful of these common traps, you can approach problems on forming quadratic equations with greater confidence and accuracy. Always take a moment to double-check signs, implied conditions (like real coefficients), and calculations.

⭐ Key Takeaways

The formation of a quadratic equation when its roots are known is a fundamental concept in Quadratic Equations. Mastering this not only helps in direct questions but also forms a basis for higher-order polynomial formation and problems involving root transformations.



Key Takeaways: Forming Quadratic Equations from Roots



  • Fundamental Principle: If $alpha$ and $eta$ are the roots of a quadratic equation, then the equation can be written as:


    $x^2 - (alpha + eta)x + (alpha eta) = 0$


    This form assumes the leading coefficient is 1. If a specific leading coefficient '$a$' is required, the equation becomes $a[x^2 - (alpha + eta)x + (alpha eta)] = 0$.

  • Steps to Form the Equation:

    1. Identify the Roots: Let the given roots be $alpha$ and $eta$.

    2. Calculate the Sum of Roots: $S = alpha + eta$.

    3. Calculate the Product of Roots: $P = alpha eta$.

    4. Substitute into the Formula: $x^2 - Sx + P = 0$.



  • Relationship with Coefficients:
    For a quadratic equation $ax^2 + bx + c = 0$ ($a
    eq 0$):

    • Sum of roots ($alpha + eta$) = $-b/a$

    • Product of roots ($alpha eta$) = $c/a$


    The formula for formation directly uses these relationships, setting $a=1$ for simplicity initially.

  • Conjugate Roots (JEE Emphasis):

    This is a crucial concept for JEE. If the coefficients of a quadratic equation are real, then:



    • If one root is irrational (e.g., $p + sqrt{q}$ where $sqrt{q}$ is irrational), then the other root must be its conjugate ($p - sqrt{q}$).

    • If one root is complex (e.g., $p + iq$ where $i = sqrt{-1}$), then the other root must be its complex conjugate ($p - iq$).


    This property is extremely useful because if you are given only one irrational or complex root, you automatically know the other root, allowing you to find the sum and product easily without needing additional information.



  • CBSE vs. JEE Perspective:

    • CBSE: Primarily focuses on applying the basic formula for real, rational, or sometimes simple irrational roots.

    • JEE: Extends to more complex scenarios including complex roots, irrational roots, and problems where the roots of one equation are used to form another equation (e.g., roots are squares or reciprocals of the original roots). The conjugate root property is frequently tested.



  • Example Quick Check:
    If roots are $2$ and $3$:
    Sum $S = 2+3 = 5$
    Product $P = 2 imes 3 = 6$
    Equation: $x^2 - 5x + 6 = 0$

  • Always Verify: After forming the equation, you can quickly verify by solving it to see if you get the original roots, or by checking if the sum and product of the roots match your calculated values.


Understanding these points provides a strong foundation for tackling various problems related to quadratic equations.

🧩 Problem Solving Approach

Forming a quadratic equation when its roots are given is a fundamental concept in Quadratic Equations. This section outlines the most efficient problem-solving approaches for such scenarios.



Method 1: Using the Sum and Product of Roots (JEE Preferred)


This is generally the most straightforward and fastest method, particularly useful when dealing with complex or irrational roots. A quadratic equation with roots α (alpha) and β (beta) can be expressed as:


x² - (Sum of Roots)x + (Product of Roots) = 0


Or, more formally:


x² - (α + β)x + (αβ) = 0



Problem-Solving Steps:



  1. Identify the given roots: Let the two given roots be α and β.

  2. Calculate the Sum of Roots: Find S = α + β.

  3. Calculate the Product of Roots: Find P = αβ.

  4. Substitute into the formula: Replace S and P into the general form x² - Sx + P = 0.

  5. Simplify: Ensure the coefficients are in their simplest integer form (if the problem doesn't specify non-integer coefficients).



Method 2: Using the Factor Form


If α and β are the roots of a quadratic equation, then (x - α) and (x - β) must be its factors. Therefore, the equation can be written as:


k(x - α)(x - β) = 0


where 'k' is any non-zero constant. Unless a specific condition (like the coefficient of x² or a point the graph passes through) is given, we generally take k = 1 for the simplest form of the equation.



Problem-Solving Steps:



  1. Identify the given roots: Let the two given roots be α and β.

  2. Form the factors: Write them as (x - α) and (x - β).

  3. Multiply the factors: Expand the expression (x - α)(x - β). This will result in a quadratic expression.

  4. Set to zero: Equate the resulting quadratic expression to zero.

  5. Consider 'k' (if applicable): If the problem specifies that the coefficient of x² should be a particular value, or if the equation passes through a specific point, use that information to determine the value of 'k'. Otherwise, assume k = 1.



JEE vs. CBSE Insight:



  • For both CBSE Board Exams and JEE Main, both methods are valid.

  • However, for JEE Main, the Sum and Product method is almost always faster and less prone to algebraic errors, especially when roots involve complex numbers or surds. It directly gives the coefficients.



Key Tips for Efficient Problem Solving:



  • Conjugate Roots: If one root is a + ib (complex) or a + &sqrt;b (irrational), and the coefficients of the quadratic equation are real/rational respectively, then its conjugate (a - ib or a - &sqrt;b) must also be a root. (Be careful: this property holds if coefficients are real/rational respectively. If not, the roots don't have to be conjugates.)

  • Simplify early: Calculate the sum and product carefully before substituting into the formula. This minimizes errors.

  • Verification: If time permits, you can quickly check your formed equation by substituting one of the given roots back into it to see if it satisfies the equation.



Example Problem:


Form a quadratic equation whose roots are 2 + &sqrt;3 and 2 - &sqrt;3.



Solution (Using Sum and Product Method):



  1. Given roots: α = 2 + &sqrt;3, β = 2 - &sqrt;3.

  2. Sum of Roots (S):

    S = α + β = (2 + &sqrt;3) + (2 - &sqrt;3) = 2 + 2 = 4


  3. Product of Roots (P):

    P = αβ = (2 + &sqrt;3)(2 - &sqrt;3)

    Using (a + b)(a - b) = a² - b²,

    P = 2² - (&sqrt;3)² = 4 - 3 = 1


  4. Substitute into formula:

    x² - Sx + P = 0

    x² - (4)x + (1) = 0

    x² - 4x + 1 = 0



The required quadratic equation is x² - 4x + 1 = 0.

📝 CBSE Focus Areas

CBSE Focus Areas: Formation of Quadratic Equations with Given Roots



For CBSE Board Examinations, the topic of 'Formation of Quadratic Equations with Given Roots' is fundamental and frequently tested. The emphasis is on understanding the core relationship between roots and coefficients and applying standard formulas accurately. Unlike JEE, CBSE typically focuses on direct application and straightforward manipulation, rather than complex or highly abstract scenarios.



Core Concept & Formula


The most crucial aspect for CBSE is the direct formula for forming a quadratic equation when its roots are known. If $alpha$ and $eta$ are the roots of a quadratic equation, the equation can be formed as:


x² - (sum of roots)x + (product of roots) = 0


Which is often written as:


x² - (α + β)x + αβ = 0


CBSE questions will primarily test your ability to correctly identify the sum (α + β) and product (αβ) of the given roots and substitute them into this formula.



Key Areas of Focus for CBSE:




  • Direct Application: You must be proficient in forming an equation when numerical roots (e.g., 2 and 3, or -1/2 and 4) are directly provided. This is the most basic and common type of question.


  • Vieta's Formulas: While not explicitly forming the equation, a strong understanding of Vieta's formulas (sum of roots = -b/a, product of roots = c/a for $ax^2 + bx + c = 0$) is foundational. CBSE often tests this by providing one equation and asking to form another whose roots are related to the first.


  • Equations with Transformed Roots: A common CBSE question type involves finding a new quadratic equation whose roots are a transformation of the roots of a given quadratic equation.


    For example, if $alpha, eta$ are roots of $ax^2 + bx + c = 0$, form an equation whose roots are:

    • $alpha^2, eta^2$

    • $1/alpha, 1/eta$

    • $alpha+k, eta+k$ or $kalpha, keta$


    In such cases, first find (α + β) and (αβ) from the given equation. Then calculate the sum and product of the *new* roots in terms of (α + β) and (αβ), and finally substitute into the standard formula.


  • Conjugate Roots Principle:

    • If the coefficients of a quadratic equation are rational, and one root is irrational (e.g., $a + sqrt{b}$), then the other root must be its conjugate ($a - sqrt{b}$).

    • If the coefficients of a quadratic equation are real, and one root is complex (e.g., $a + ib$), then the other root must be its conjugate ($a - ib$).


    This principle is very important for CBSE and often forms the basis of problems where only one root is provided, and you need to infer the other.



Example (CBSE-Style):


Question: If $alpha, eta$ are the roots of the equation $2x^2 - 3x + 1 = 0$, form a quadratic equation whose roots are $1/alpha$ and $1/eta$.


Solution:
From $2x^2 - 3x + 1 = 0$, we have:
Sum of roots, $alpha + eta = -(-3)/2 = 3/2$
Product of roots, $alphaeta = 1/2$


For the new equation, the roots are $1/alpha$ and $1/eta$.



  • New Sum of roots = $1/alpha + 1/eta = (alpha + eta) / (alphaeta) = (3/2) / (1/2) = 3$

  • New Product of roots = $(1/alpha) imes (1/eta) = 1/(alphaeta) = 1/(1/2) = 2$


The required quadratic equation is $x^2 - ( ext{New Sum})x + ( ext{New Product}) = 0$
$x^2 - 3x + 2 = 0$




CBSE Tip: Always write down the general formula $x^2 - (S)x + P = 0$ first. This helps in organizing your solution and earns step marks even if there's a minor calculation error. Pay close attention to signs!


🎓 JEE Focus Areas

The ability to form a quadratic equation when its roots are known is fundamental. However, JEE Main often tests this concept with added layers of complexity, requiring a deeper understanding of root properties and their relationships with coefficients.



The basic formula for forming a quadratic equation with roots α and β is:


x2 – (Sum of Roots)x + (Product of Roots) = 0


Or, x2 – (α + β)x + αβ = 0



JEE Focus Areas:




  • Conjugate Roots (Critical for JEE):

    • If a quadratic equation has real coefficients, and one root is complex (a + ib), then the other root *must* be its conjugate (a - ib).

    • If a quadratic equation has rational coefficients, and one root is irrational (a + √b), then the other root *must* be its conjugate (a - √b).

    • JEE Trap: This rule holds ONLY if coefficients are real/rational respectively. If coefficients are themselves complex or irrational, the conjugate property may not apply. For example, if the equation is $x^2 - (2+i)x + (1+i) = 0$, and one root is $1+i$, the other root is not necessarily $1-i$.




  • Transformation of Equations:
    Often, questions involve finding a new quadratic equation whose roots are functions of the roots of a *given* quadratic equation.

    • If α, β are roots of $ax^2+bx+c=0$, then $α + β = -b/a$ and $αβ = c/a$.

    • To find an equation with roots like $1/α, 1/β$ or $α^2, β^2$ or $α+k, β+k$:

      1. Calculate the sum of the new roots using the relationships derived from the original equation.

      2. Calculate the product of the new roots similarly.

      3. Substitute these new sum and product values into the standard formula $x^2 - (S_{new})x + (P_{new}) = 0$.






  • Symmetric Expressions of Roots:
    Many JEE problems require expressing $alpha^2+eta^2$, $alpha^3+eta^3$, etc., in terms of $alpha+eta$ and $alphaeta$. Familiarity with these identities is crucial.

    • $alpha^2+eta^2 = (alpha+eta)^2 - 2alphaeta$

    • $alpha^3+eta^3 = (alpha+eta)(alpha^2-alphaeta+eta^2) = (alpha+eta)[(alpha+eta)^2-3alphaeta]$

    • $alpha/eta + eta/alpha = (alpha^2+eta^2)/(alphaeta)$




  • Identifying Roots from Coefficient Relations:
    Sometimes, conditions on roots are given (e.g., one root is $k$ times the other, roots differ by $k$). These conditions can be used in conjunction with Vieta's formulas to establish relationships between the coefficients, indirectly leading to forming or identifying properties of quadratic equations.



CBSE vs. JEE Main:






















Aspect CBSE Board Level JEE Main Level
Complexity Direct application of formula; simple conjugate roots. Involves transformation of roots, using symmetric expressions, and understanding conditions for conjugate roots with non-real/non-rational coefficients.
Roots Given Simple integers, fractions, or direct conjugates. Complex numbers, surds, or roots expressed as functions of other roots.


Example:

If α, β are the roots of the equation $x^2 - 4x + 2 = 0$, form the quadratic equation whose roots are $1/alpha$ and $1/eta$.


Solution:

For $x^2 - 4x + 2 = 0$, we have $alpha + eta = -(-4)/1 = 4$ and $alphaeta = 2/1 = 2$.


For the new equation, let the roots be $alpha'$ and $eta'$. Here, $alpha' = 1/alpha$ and $eta' = 1/eta$.



  • Sum of new roots:

    $alpha' + eta' = 1/alpha + 1/eta = (alpha+eta)/(alphaeta) = 4/2 = 2$

  • Product of new roots:

    $alpha' eta' = (1/alpha)(1/eta) = 1/(alphaeta) = 1/2$


The new quadratic equation is $x^2 - (alpha'+eta')x + (alpha'eta') = 0$

$x^2 - (2)x + (1/2) = 0$

Multiplying by 2 to clear the fraction: $2x^2 - 4x + 1 = 0$.


Mastering these nuances will significantly boost your performance in JEE Main questions related to quadratic equations.

📊 AI Generation Metadata

AI Model: Gemini Full Parallel API Client
Status: AI Generated
Version: 1
Generated: Oct 22, 2025
🌐 Overview
Goal: build a quadratic equation from specified roots. If roots are p and q, the monic quadratic is x^2 − (p + q)x + pq = 0. With a chosen leading coefficient a ≠ 0, the general form is a x^2 − a(p + q)x + a p q = 0. For real coefficients, non‑real complex or irrational surds appear in conjugate pairs. Special cases include repeated roots (p = q), reciprocal/negated/transformed roots, and constraints given via sum S and product P (then equation is x^2 − Sx + P = 0).

Micro‑examples:
• Roots 2 and −3 ⇒ x^2 + x − 6 = 0.
• Root 2 + i with real coefficients ⇒ other root 2 − i, so x^2 − 4x + 5 = 0.
• Roots ±√3 ⇒ x^2 − 3 = 0.
📚 Fundamentals
• Base formula: roots p, q ⇒ x^2 − (p + q)x + pq = 0 (monic).
• General a ≠ 0: a x^2 − a(p + q)x + a p q = 0.
• Conjugate rule (real coefficients): if p is non‑real or an irrational simple surd, its conjugate must also be a root.
• Transforms: for roots kα and kβ ⇒ S′ = k(α + β), P′ = k^2 αβ; for α + k, β + k ⇒ S′ = (α + β) + 2k, P′ = αβ + k(α + β) + k^2; for 1/α, 1/β (αβ ≠ 0) ⇒ S′ = (α + β)/(αβ), P′ = 1/(αβ).
• Repeated root r ⇒ (x − r)^2 = 0.
🔬 Deep Dive
Vieta's formulas generalize formation: coefficients are (up to sign) the elementary symmetric sums of roots. For quadratics, c/a = αβ and −b/a = α + β. The minimal‑polynomial idea explains why conjugate roots appear for real coefficients: complex conjugation is a field automorphism preserving real coefficients, so conjugates share the same polynomial. Formation with constraints often reduces to computing S, P from conditions and building x^2 − Sx + P = 0, then scaling.
🎯 Shortcuts
• "Minus sum, plus product": x^2 − Sx + P.
• "Conjugate comes home": for real coefficients, bring the conjugate.
• "Clear to integer": multiply through to remove fractions.
💡 Quick Tips
• Always verify with Vieta: sum → coefficient of x, product → constant term.
• Keep track of the sign on S carefully.
• If a is given, multiply every term; do not alter S or P before scaling.
• For surds, prefer rationalized/integer coefficients by clearing denominators.
• If only one root is provided and coefficients must be real, deduce the other by conjugation.
🧠 Intuitive Understanding
Forming an equation is like "packaging" two numbers into a single object whose DNA (coefficients) stores their sum and product. Multiplying factors (x − p)(x − q) expands to coefficients that encode p + q and pq. If coefficients are real and one root is complex or a simple surd, its mirror (conjugate) must be present so the packaging remains real.
🌍 Real World Applications
• Constructing models from desired equilibria or steady states (roots as setpoints).
• Designing polynomials with specified spectral properties in control/filters (quadratic sections).
• Geometry: constructing circles/lines intersection conditions that reverse‑engineer required roots.
• Problem setting: from given sum/product/transform of roots, write the governing quadratic quickly.
🔄 Common Analogies
• Recipe card: ingredients p and q produce coefficients S = p + q and P = pq.
• Lock and key: the factor (x − p)(x − q) is the lock built for keys p and q.
• Mirror rule: for real coefficients, non‑real roots must come with their mirror image (conjugate).
📋 Prerequisites
• Expanding products and comparing coefficients.
• Vieta's relations for quadratics: S = α + β = −b/a, P = αβ = c/a.
• Complex conjugates and irrational conjugates with real coefficients.
• Algebra with radicals and simple transformations of numbers.
🔗 Related Topics
Relations between roots and coefficients; nature of roots via discriminant; transformations of roots (kα, α + k, 1/α, −α); reciprocal and conjugate quadratics; parameterized quadratics.
⚠️ Common Exam Traps
• Forgetting the conjugate root for real coefficients.
• Sign error in the middle coefficient (should be −S).
• Leaving fractional coefficients when an integer‑coefficient form is expected.
• Misapplying transforms (e.g., using S and P before applying a scale/shift).
• Ignoring a specified leading coefficient a and writing only the monic form.
⭐ Key Takeaways
• Coefficients encode sum and product of roots.
• Choose monic form unless a specific leading coefficient is required.
• Ensure conjugate pairs for real‑coefficient quadratics.
• For transformed roots, recompute S and P first, then write x^2 − Sx + P = 0.
• Clear denominators to present neat integer coefficients when possible.
🧩 Problem Solving Approach
Algorithm: (1) Identify roots or their relations (sum/product or transforms). (2) Compute S = p + q and P = pq (or S′, P′ after transforms). (3) Write x^2 − Sx + P = 0; multiply by a if leading coefficient is specified. (4) If coefficients must be real and a given root is complex/irrational, include its conjugate. Example: Roots 3 and 1/2 ⇒ S = 3.5, P = 1.5 ⇒ 2x^2 − 7x + 3 = 0 (after clearing denominators).
📝 CBSE Focus Areas
• Writing x^2 − Sx + P = 0 from given roots.
• Handling repeated roots and simple conjugate cases.
• Clearing fractions to present clean integer coefficients.
• Short numericals verifying with Vieta's relations.
🎓 JEE Focus Areas
• Transformed roots (kα, α + k, 1/α, −α) and parameterized formation.
• Real‑coefficient constraints forcing conjugate pairs.
• Minimal polynomial viewpoint and coefficient integrality conditions.
• Mixed problems combining nature/relations/formation in one.

📊 AI Generation Metadata

Difficulty: Intermediate
Estimated Time: 90 mins
AI Model: ChatGPT 5 (Copilot Agent)
Status: AI Generated
Version: 1
Generated: Oct 23, 2025
🌐 Overview
Forming a quadratic equation given its roots is the inverse of solving a quadratic. It combines Vieta's formulas, polynomial construction, and algebraic manipulation. This skill is crucial for CBSE and IIT-JEE, enabling rapid problem-solving in optimization, symmetry analysis, and polynomial relationships.
📚 Fundamentals
Vieta's Formulas (Fundamental Relationships):
For quadratic ( ax^2 + bx + c = 0 ) with roots α and β:
Sum of roots: ( alpha + eta = -frac{b}{a} )
Product of roots: ( alpha eta = frac{c}{a} )

Formation of Quadratic Given Two Roots:

Method 1: Using Sum and Product
If roots are α and β, the quadratic is:
( x^2 - (alpha + eta)x + alphaeta = 0 )
or in standard form:
( x^2 - (S)x + (P) = 0 )
where S = α + β, P = αβ

Example 1: Roots 2 and 3
Sum = 5, Product = 6
Quadratic: ( x^2 - 5x + 6 = 0 )

Example 2: Roots ( 3 + sqrt{2} ) and ( 3 - sqrt{2} )
Sum = 6, Product = (3)² - (√2)² = 9 - 2 = 7
Quadratic: ( x^2 - 6x + 7 = 0 )

Method 2: Using Linear Factors
If roots are α and β, the quadratic is:
( (x - alpha)(x - eta) = 0 )
Expanding: ( x^2 - (alpha + eta)x + alphaeta = 0 )

Example: Roots are complex conjugates ( 2 + 3i ) and ( 2 - 3i )
( [(x - (2 + 3i))][(x - (2 - 3i))] = 0 )
( [(x - 2) - 3i][(x - 2) + 3i] = 0 )
( [(x - 2)]^2 - (3i)^2 = 0 )
( (x - 2)^2 + 9 = 0 )
( x^2 - 4x + 4 + 9 = 0 )
( x^2 - 4x + 13 = 0 )

Scaling: Quadratic with Coefficient a
If roots are α and β and leading coefficient is a:
( a(x - alpha)(x - eta) = 0 )
( ax^2 - a(alpha + eta)x + aalphaeta = 0 )

Example: Roots 1/2, 1/3; leading coefficient 6
( 6(x - 1/2)(x - 1/3) = 0 )
( 6x^2 - 6(1/2 + 1/3)x + 6(1/2)(1/3) = 0 )
( 6x^2 - 6(5/6)x + 6(1/6) = 0 )
( 6x^2 - 5x + 1 = 0 )
🔬 Deep Dive
Advanced Transformations:

1. Equation with Roots (α + k), (β + k) [Shifted Roots]:
If α, β are roots of ( f(x) = x^2 + px + q = 0 )
Then (α + k), (β + k) are roots of:
( f(x - k) = (x - k)^2 + p(x - k) + q = 0 )
( x^2 - 2kx + k^2 + px - pk + q = 0 )
( x^2 + (p - 2k)x + (k^2 - pk + q) = 0 )

Shortcut: Replace x with (x - k) in original equation.

Example: ( x^2 - 5x + 6 = 0 ) has roots 2, 3.
Equation with roots 4, 5: Replace x → (x - 2)
( (x - 2)^2 - 5(x - 2) + 6 = 0 )
( x^2 - 4x + 4 - 5x + 10 + 6 = 0 )
( x^2 - 9x + 20 = 0 )
Verify: Roots = (9 ± √(81 - 80))/2 = (9 ± 1)/2 = 5, 4 ✓

2. Equation with Roots kα, kβ [Scaled Roots]:
If α, β are roots of ( x^2 + px + q = 0 )
Then kα, kβ are roots of:
( (x/k)^2 + p(x/k) + q = 0 )
( x^2/k^2 + px/k + q = 0 )
Multiply by k²:
( x^2 + pkx + qk^2 = 0 )

Shortcut: Replace x with (x/k) and clear fractions.

Example: ( x^2 - 5x + 6 = 0 ) has roots 2, 3.
Equation with roots 4, 6 (doubled): Replace x → (x/2)
( (x/2)^2 - 5(x/2) + 6 = 0 )
Multiply by 4:
( x^2 - 10x + 24 = 0 )
Verify: Roots = (10 ± √(100 - 96))/2 = (10 ± 2)/2 = 6, 4 ✓

3. Equation with Roots 1/α, 1/β [Reciprocal Roots]:
If α, β are roots of ( ax^2 + bx + c = 0 )
Then 1/α, 1/β are roots of:
( a(1/x)^2 + b(1/x) + c = 0 )
( a/x^2 + b/x + c = 0 )
Multiply by x²:
( a + bx + cx^2 = 0 )
( cx^2 + bx + a = 0 ) (coefficients reversed)

Example: ( x^2 - 5x + 6 = 0 ) has roots 2, 3.
Equation with roots 1/2, 1/3:
( 6x^2 - 5x + 1 = 0 )
Verify: x = (5 ± √(25 - 24))/12 = (5 ± 1)/12 = 1/2, 1/3 ✓

4. Equation with Roots α², β² [Squared Roots]:
Sum: ( alpha^2 + eta^2 = (alpha + eta)^2 - 2alphaeta = p^2 - 2q )
Product: ( alpha^2 eta^2 = (alphaeta)^2 = q^2 )
Equation: ( x^2 - (p^2 - 2q)x + q^2 = 0 )

Example: ( x^2 - 5x + 6 = 0 ) has roots 2, 3.
Equation with roots 4, 9:
( x^2 - (25 - 12)x + 36 = 0 )
( x^2 - 13x + 36 = 0 )
Verify: Roots = (13 ± √(169 - 144))/2 = (13 ± 5)/2 = 9, 4 ✓

5. Combined Conditions (IIT-JEE):
"Form equation with roots α² + β², α²β²":
Sum: ( alpha^2 + eta^2 = p^2 - 2q ) (from above)
Product: ( (alpha^2 + eta^2)(alpha^2eta^2) = (p^2 - 2q)q^2 )
Hmm, this is not a simple sum-product of two roots directly; requires further analysis.

Better approach: Define new variables u = α² + β², v = α²β² = q²
We need u + v and uv to form an equation in these:
This becomes complex; usually approached by solving iteratively or using root properties.
🎯 Shortcuts
"Sum and product → equation: x² - Sx + P = 0." "Complex conjugates → real coefficients." "Reciprocals → coefficients reversed." "Shift x → (x-k), Scale x → (x/k)."
💡 Quick Tips
Always work with monic polynomial (leading coefficient 1) for sum-product method, then scale if needed. For irrational/complex roots, use conjugate pairs to ensure real coefficients. When verifying, substitute one root back into your equation to check.
🧠 Intuitive Understanding
You have two "puzzle pieces" (the roots), and you need to reconstruct the "box" (the equation) they came from. Using Vieta's secrets (sum and product), you build the equation backward.
🌍 Real World Applications
Engineering: designing systems with specific resonance frequencies. Finance: modeling growth with prescribed rates. Physics: constructing potential energy equations from eigenvalues.
🔄 Common Analogies
Forming a quadratic from roots is like reconstructing a parabola from its x-intercepts. The Vieta's formulas tell you the "distance between intercepts" (related to sum) and their "product location" on the axes.Forming a quadratic from roots is like reconstructing a parabola from its x-intercepts. The Vieta's formulas tell you the "distance between intercepts" (related to sum) and their "product location" on the axes.
📋 Prerequisites
Quadratic equations, Vieta's formulas, complex numbers basics, algebraic manipulation.
🔗 Related Topics
Vieta's Formulas, Quadratic Equations, Relations Between Roots and Coefficients, Complex Numbers, Polynomials
⚠️ Common Exam Traps
Forgetting negative sign in ( -(alpha + eta) ) for sum term. Not recognizing complex conjugates (giving real coefficients). Incorrect transformation application (e.g., confusing shift direction). Not verifying formed equation by substitution. Arithmetic errors in sum/product calculation.
⭐ Key Takeaways
Fundamental: ( x^2 - ( ext{sum})x + ext{product} = 0 ). Complex conjugates ( p ± qi ) give real coefficients: ( x^2 - 2px + (p^2 + q^2) = 0 ). Reciprocal roots swap coefficients. Transformations: shift via x → (x - k), scale via x → (x/k).
🧩 Problem Solving Approach
Step 1: Identify given roots. Step 2: Calculate sum S = α + β and product P = αβ. Step 3: Form equation: ( x^2 - Sx + P = 0 ). Step 4: If transformation (shifted, scaled, etc.), identify the transformation rule. Step 5: Apply transformation and simplify. Step 6: Verify by solving resulting equation.
📝 CBSE Focus Areas
Forming equations given two roots. Using sum and product (Vieta's). Handling irrational roots. Complex conjugate pairs. Standard quadratic forms. Verification by solving.
🎓 JEE Focus Areas
Advanced transformations (shifted, scaled, reciprocal, squared roots). Parametric equations with root constraints. Combined conditions (e.g., roots differing by k, sum/product constraints). Multi-step formation problems. Complex root formations. Equations with functional root relationships.

📊 AI Generation Metadata

Difficulty: Beginner
Estimated Time: 35 mins
AI Model: Claude Haiku 4.5 (Preview)
Status: AI Generated
Version: 1
Generated: Oct 18, 2025

📝CBSE 12th Board Problems (12)

Problem 255
Easy 2 Marks
Find the quadratic equation whose roots are 3 and 5.
Show Solution
1. Calculate the sum of the roots: S = α + β = 3 + 5 = 8. 2. Calculate the product of the roots: P = α * β = 3 * 5 = 15. 3. Use the formula for a quadratic equation: x² - (Sum of roots)x + (Product of roots) = 0. 4. Substitute the values: x² - 8x + 15 = 0.
Final Answer: x² - 8x + 15 = 0
Problem 255
Easy 2 Marks
Form a quadratic equation if its roots are -2 and 7.
Show Solution
1. Sum of roots: S = -2 + 7 = 5. 2. Product of roots: P = (-2) * 7 = -14. 3. Equation: x² - Sx + P = 0. 4. Substitute: x² - 5x + (-14) = 0 => x² - 5x - 14 = 0.
Final Answer: x² - 5x - 14 = 0
Problem 255
Easy 3 Marks
Construct a quadratic equation whose roots are 1/2 and 4.
Show Solution
1. Sum of roots: S = 1/2 + 4 = 1/2 + 8/2 = 9/2. 2. Product of roots: P = (1/2) * 4 = 2. 3. Equation: x² - Sx + P = 0. 4. Substitute: x² - (9/2)x + 2 = 0. 5. Multiply by 2 to clear the fraction: 2x² - 9x + 4 = 0.
Final Answer: 2x² - 9x + 4 = 0
Problem 255
Easy 2 Marks
Write the quadratic equation with roots -3 and -6.
Show Solution
1. Sum of roots: S = -3 + (-6) = -9. 2. Product of roots: P = (-3) * (-6) = 18. 3. Equation: x² - Sx + P = 0. 4. Substitute: x² - (-9)x + 18 = 0 => x² + 9x + 18 = 0.
Final Answer: x² + 9x + 18 = 0
Problem 255
Easy 2 Marks
Determine the quadratic equation whose roots are 0 and 9.
Show Solution
1. Sum of roots: S = 0 + 9 = 9. 2. Product of roots: P = 0 * 9 = 0. 3. Equation: x² - Sx + P = 0. 4. Substitute: x² - 9x + 0 = 0 => x² - 9x = 0.
Final Answer: x² - 9x = 0
Problem 255
Easy 2 Marks
Find the quadratic equation if the sum of its roots is 4 and the product of its roots is -12.
Show Solution
1. The general form of a quadratic equation is x² - (Sum of roots)x + (Product of roots) = 0. 2. Substitute the given sum and product directly into the formula. 3. x² - (4)x + (-12) = 0. 4. Simplify: x² - 4x - 12 = 0.
Final Answer: x² - 4x - 12 = 0
Problem 255
Medium 2 Marks
Form a quadratic equation whose roots are 3 and -5.
Show Solution
1. Identify the given roots, let α = 3 and β = -5. 2. Calculate the sum of the roots: α + β = 3 + (-5) = -2. 3. Calculate the product of the roots: αβ = 3 * (-5) = -15. 4. Use the formula for forming a quadratic equation: x² - (α + β)x + αβ = 0. 5. Substitute the sum and product into the formula: x² - (-2)x + (-15) = 0. 6. Simplify the equation.
Final Answer: x² + 2x - 15 = 0
Problem 255
Medium 2 Marks
Find the quadratic equation whose roots are 1/2 and 2/3.
Show Solution
1. Let α = 1/2 and β = 2/3. 2. Calculate the sum of roots: α + β = 1/2 + 2/3 = 3/6 + 4/6 = 7/6. 3. Calculate the product of roots: αβ = (1/2) * (2/3) = 2/6 = 1/3. 4. Apply the formula: x² - (α + β)x + αβ = 0. 5. Substitute the values: x² - (7/6)x + 1/3 = 0. 6. Multiply the entire equation by the LCM of denominators (6) to clear fractions.
Final Answer: 6x² - 7x + 2 = 0
Problem 255
Medium 3 Marks
Construct a quadratic equation having roots 2 + √5 and 2 - √5.
Show Solution
1. Let α = 2 + √5 and β = 2 - √5. 2. Calculate the sum of roots: α + β = (2 + √5) + (2 - √5) = 4. 3. Calculate the product of roots: αβ = (2 + √5)(2 - √5). Use the identity (a+b)(a-b) = a² - b². 4. αβ = 2² - (√5)² = 4 - 5 = -1. 5. Apply the formula: x² - (α + β)x + αβ = 0. 6. Substitute the values: x² - (4)x + (-1) = 0.
Final Answer: x² - 4x - 1 = 0
Problem 255
Medium 2 Marks
Form the quadratic equation whose roots are -2 and 1/3.
Show Solution
1. Let α = -2 and β = 1/3. 2. Calculate the sum of roots: α + β = -2 + 1/3 = -6/3 + 1/3 = -5/3. 3. Calculate the product of roots: αβ = (-2) * (1/3) = -2/3. 4. Apply the formula: x² - (α + β)x + αβ = 0. 5. Substitute the values: x² - (-5/3)x + (-2/3) = 0. 6. Multiply the entire equation by 3 to clear fractions.
Final Answer: 3x² + 5x - 2 = 0
Problem 255
Medium 2 Marks
Determine the quadratic equation with roots (3/4) and (1/5).
Show Solution
1. Let α = 3/4 and β = 1/5. 2. Calculate the sum of roots: α + β = 3/4 + 1/5 = 15/20 + 4/20 = 19/20. 3. Calculate the product of roots: αβ = (3/4) * (1/5) = 3/20. 4. Apply the formula: x² - (α + β)x + αβ = 0. 5. Substitute the values: x² - (19/20)x + 3/20 = 0. 6. Multiply the entire equation by 20 to clear fractions.
Final Answer: 20x² - 19x + 3 = 0
Problem 255
Medium 3 Marks
Form a quadratic equation with real coefficients if one of its roots is 1 + i, where 'i' is the imaginary unit.
Show Solution
1. If one root of a quadratic equation with real coefficients is complex (a+bi), then its conjugate (a-bi) must also be a root. So, α = 1 + i and β = 1 - i. 2. Calculate the sum of roots: α + β = (1 + i) + (1 - i) = 2. 3. Calculate the product of roots: αβ = (1 + i)(1 - i). Use the identity (a+b)(a-b) = a² - b². 4. αβ = 1² - i² = 1 - (-1) = 1 + 1 = 2. 5. Apply the formula: x² - (α + β)x + αβ = 0. 6. Substitute the values: x² - (2)x + 2 = 0.
Final Answer: x² - 2x + 2 = 0

🎯IIT-JEE Main Problems (18)

Problem 255
Medium 4 Marks
If α and β are the roots of the equation x² - x - 1 = 0, form the quadratic equation whose roots are α⁸ and β⁸.
Show Solution
1. From the given equation, α + β = 1 and αβ = -1. 2. We need to find α⁸ and β⁸. Since α and β are roots of x² - x - 1 = 0, they satisfy the equation: α² - α - 1 = 0 => α² = α + 1. 3. Similarly, β² = β + 1. 4. Calculate higher powers of α: α³ = α² + α = (α + 1) + α = 2α + 1. α⁴ = 2α² + α = 2(α + 1) + α = 3α + 2. α⁸ = (α⁴)² = (3α + 2)² = 9α² + 12α + 4. 5. Substitute α² = α + 1 into α⁸ expression: α⁸ = 9(α + 1) + 12α + 4 = 9α + 9 + 12α + 4 = 21α + 13. 6. Similarly, for β: β⁸ = 21β + 13. 7. Calculate the sum of the new roots: S' = α⁸ + β⁸ = (21α + 13) + (21β + 13) = 21(α + β) + 26. 8. Substitute α + β = 1: S' = 21(1) + 26 = 47. 9. Calculate the product of the new roots: P' = α⁸β⁸ = (αβ)⁸. 10. Substitute αβ = -1: P' = (-1)⁸ = 1. 11. The required quadratic equation is x² - S'x + P' = 0. 12. Substitute S' = 47 and P' = 1: x² - 47x + 1 = 0.
Final Answer: x² - 47x + 1 = 0
Problem 255
Hard 4 Marks
If α and β are the roots of the quadratic equation x² - x - 1 = 0, form a quadratic equation whose roots are (α^8 + β^8) and (α^9 + β^9).
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1. For the given equation x² - x - 1 = 0: α + β = 1 αβ = -1 2. Let S_n = α^n + β^n. We need to find S_8 and S_9. Since α and β are roots of x² - x - 1 = 0, they satisfy the recurrence relation: S_n = S_(n-1) + S_(n-2) for n ≥ 2. 3. Calculate initial values: S_0 = α^0 + β^0 = 1 + 1 = 2. S_1 = α + β = 1. 4. Calculate S_n iteratively: S_2 = S_1 + S_0 = 1 + 2 = 3. S_3 = S_2 + S_1 = 3 + 1 = 4. S_4 = S_3 + S_2 = 4 + 3 = 7. S_5 = S_4 + S_3 = 7 + 4 = 11. S_6 = S_5 + S_4 = 11 + 7 = 18. S_7 = S_6 + S_5 = 18 + 11 = 29. S_8 = S_7 + S_6 = 29 + 18 = 47. S_9 = S_8 + S_7 = 47 + 29 = 76. 5. The new roots are R&sub1; = S_8 = 47 and R&sub2; = S_9 = 76. 6. For the new quadratic equation: Sum of new roots (S') = R&sub1; + R&sub2; = 47 + 76 = 123. Product of new roots (P') = R&sub1; * R&sub2; = 47 * 76 = 3572. 7. The new quadratic equation is x² - S'x + P' = 0. x² - 123x + 3572 = 0.
Final Answer: x² - 123x + 3572 = 0
Problem 255
Hard 4 Marks
If α and β are the roots of the quadratic equation x² - 6x + a = 0 and satisfy the relation 3α + 2β = 20, form a quadratic equation whose roots are 'a' and '(a+1)'.
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1. For the given equation x² - 6x + a = 0: Sum of roots (α + β) = 6. Product of roots (αβ) = a. 2. We are given a system of equations involving α and β: (i) α + β = 6 (ii) 3α + 2β = 20 3. From (i), β = 6 - α. Substitute this into (ii): 3α + 2(6 - α) = 20 3α + 12 - 2α = 20 α + 12 = 20 α = 8. 4. Now find β using β = 6 - α: β = 6 - 8 = -2. 5. Using the product of roots αβ = a: (8)(-2) = a a = -16. 6. The new roots for the quadratic equation are 'a' and '(a+1)'. Substitute the value of a = -16: New roots are -16 and (-16 + 1) = -15. 7. For the new quadratic equation: Sum of new roots (S') = -16 + (-15) = -31. Product of new roots (P') = (-16)(-15) = 240. 8. The new quadratic equation is x² - S'x + P' = 0. x² - (-31)x + 240 = 0. x² + 31x + 240 = 0.
Final Answer: x² + 31x + 240 = 0
Problem 255
Hard 4 Marks
If α and β are the roots of the quadratic equation x² - px + q = 0, form a quadratic equation whose roots are ( α² + 1/β² ) and ( β² + 1/α² ).
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1. For the given equation x² - px + q = 0: α + β = p αβ = q 2. We need to find the sum (S') and product (P') of the new roots, R&sub1; = α² + 1/β² and R&sub2; = β² + 1/α². 3. First, express α² + β² in terms of p and q: α² + β² = (α + β)² - 2αβ = p² - 2q. 4. Calculate S' = R&sub1; + R&sub2;: S' = (α² + 1/β²) + (β² + 1/α²) S' = (α² + β²) + (1/α² + 1/β²) S' = (α² + β²) + (α² + β²) / (αβ)² Substitute α² + β² = p² - 2q and αβ = q: S' = (p² - 2q) + (p² - 2q) / q² S' = (p² - 2q)(1 + 1/q²) = (p² - 2q)(q² + 1) / q². 5. Calculate P' = R&sub1; * R&sub2;: P' = (α² + 1/β²) (β² + 1/α²) P' = α²β² + α²/α² + 1/β² * β² + 1/(αβ)² P' = (αβ)² + 1 + 1 + 1/(αβ)² P' = q² + 2 + 1/q² P' = (q&sup4; + 2q² + 1) / q² = (q² + 1)² / q². 6. The new quadratic equation is x² - S'x + P' = 0. x² - [(p² - 2q)(q² + 1) / q²]x + [(q² + 1)² / q²] = 0. Multiply by q² to clear denominators: q²x² - (p² - 2q)(q² + 1)x + (q² + 1)² = 0.
Final Answer: q²x² - (p² - 2q)(q² + 1)x + (q² + 1)² = 0
Problem 255
Hard 4 Marks
If α and β are the roots of the quadratic equation x² - 5x + 1 = 0, form a quadratic equation whose roots are 2 log&sub1;₀(α) and 2 log&sub1;₀(β).
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1. For the given equation x² - 5x + 1 = 0: α + β = 5 αβ = 1 2. The new roots are R&sub1; = 2 log&sub1;₀(α) and R&sub2; = 2 log&sub1;₀(β). 3. Sum of new roots (S') = R&sub1; + R&sub2; = 2 log&sub1;₀(α) + 2 log&sub1;₀(β) S' = 2 [log&sub1;₀(α) + log&sub1;₀(β)] Using log property log a + log b = log (ab): S' = 2 log&sub1;₀(αβ). Substitute αβ = 1: S' = 2 log&sub1;₀(1). Since log&sub1;₀(1) = 0, S' = 2 * 0 = 0. 4. Product of new roots (P') = R&sub1; * R&sub2; = [2 log&sub1;₀(α)] * [2 log&sub1;₀(β)] P' = 4 [log&sub1;₀(α) log&sub1;₀(β)]. To evaluate this, we need log&sub1;₀(α) and log&sub1;₀(β) individually. Consider the original roots: x = [5 ± √(25 - 4)] / 2 = (5 ± √21) / 2. Let α = (5 + √21) / 2 and β = (5 - √21) / 2. This method is tedious. 5. Let's re-examine the product P'. Since αβ = 1, then α = 1/β. So, log&sub1;₀(α) = log&sub1;₀(1/β) = -log&sub1;₀(β). This means the new roots are R&sub1; = 2 log&sub1;₀(α) and R&sub2; = -2 log&sub1;₀(α). This implies that the new roots are additive inverses of each other (R&sub2; = -R&sub1;). For such roots, the sum S' must be 0, which we already found. Now, P' = R&sub1; * R&sub2; = [2 log&sub1;₀(α)] * [-2 log&sub1;₀(α)] = -4 [log&sub1;₀(α)]². We need to find log&sub1;₀(α). We know α + β = 5 and αβ = 1. We have the relation α + 1/α = 5. Not directly helpful for log&sub1;₀(α). Let y = log&sub1;₀(x). Then x = 10^y. The original equation is x² - 5x + 1 = 0. Consider the relation α + 1/α = 5. (This comes from dividing x² - 5x + 1 = 0 by x) This means α and β are positive real roots. √D = √(25-4) = √21 > 0. Since αβ = 1, we have β = 1/α. So the new roots are 2 log&sub1;₀(α) and 2 log&sub1;₀(1/α) = -2 log&sub1;₀(α). Let K = 2 log&sub1;₀(α). The new roots are K and -K. Sum of new roots S' = K + (-K) = 0. Product of new roots P' = K * (-K) = -K² = - [2 log&sub1;₀(α)]² = -4 [log&sub1;₀(α)]². Since α is a root of x² - 5x + 1 = 0, α = (5 + √21) / 2. P' = -4 [log&sub1;₀((5 + √21) / 2)]². This expression for P' is not typically expected in a simple integer/rational form for JEE Main. Let's check the wording of the question: 2 log&sub1;₀(α) and 2 log&sub1;₀(β). The property αβ=1 is key. log(αβ) = log(1) => log&sub1;₀(α) + log&sub1;₀(β) = 0. So, log&sub1;₀(β) = -log&sub1;₀(α). The new roots are indeed 2L and -2L where L = log&sub1;₀(α). Thus S' = 0. And P' = (2L)(-2L) = -4L² = -4(log&sub1;₀α)². This requires an explicit value for α to give a numerical answer for P'. This means the question setup probably allows for this form, or I'm missing a simplification. No, it's correct. The question asks to form the equation, not to find exact values. So the equation will be x² + 0x - 4(log&sub1;₀α)² = 0. Let's confirm if log&sub1;₀((5 + √21) / 2) can be simplified in any way. No obvious simplification. This question is hard because of the expression for P'. The equation is x² - 0x - 4(log&sub1;₀α)² = 0. So, x² - 4(log&sub1;₀α)² = 0. This is still a correct quadratic equation.
Final Answer: x² - 4[log&sub1;₀((5 + √21)/2)]² = 0 (or x² - 4[log&sub1;₀((5 - √21)/2)]² = 0, as log&sub1;₀(β) = -log&sub1;₀(α))
Problem 255
Hard 4 Marks
If α and β are the roots of the quadratic equation x² - 3x + 1 = 0, form a quadratic equation whose roots are ( α² / (β + 1) ) and ( β² / (α + 1) ).
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1. For the given equation x² - 3x + 1 = 0: α + β = 3 αβ = 1 2. Since α and β are roots, they satisfy the equation: α² - 3α + 1 = 0 ⇒ α² = 3α - 1. Similarly, β² = 3β - 1. 3. Let the new roots be R&sub1; = α² / (β + 1) and R&sub2; = β² / (α + 1). 4. Simplify R&sub1; using α² = 3α - 1: R&sub1; = (3α - 1) / (β + 1). From αβ = 1, we have β = 1/α. So, β + 1 = 1/α + 1 = (α + 1)/α. R&sub1; = (3α - 1) / [(α + 1)/α] = α(3α - 1) / (α + 1) = (3α² - α) / (α + 1). Substitute α² = 3α - 1 again: R&sub1; = (3(3α - 1) - α) / (α + 1) = (9α - 3 - α) / (α + 1) = (8α - 3) / (α + 1). This simplification is still in terms of α. Let's try to express β+1 in terms of α using α+β=3. β + 1 = (3 - α) + 1 = 4 - α. So, R&sub1; = α² / (4 - α). Substitute α² = 3α - 1: R&sub1; = (3α - 1) / (4 - α). Similarly, R&sub2; = (3β - 1) / (4 - β). 5. Sum of new roots (S') = R&sub1; + R&sub2; = (3α - 1) / (4 - α) + (3β - 1) / (4 - β). S' = [(3α - 1)(4 - β) + (3β - 1)(4 - α)] / [(4 - α)(4 - β)] Numerator: (12α - 3αβ - 4 + β) + (12β - 3αβ - 4 + α) = 13(α + β) - 6αβ - 8 = 13(3) - 6(1) - 8 = 39 - 6 - 8 = 25. Denominator: 16 - 4β - 4α + αβ = 16 - 4(α + β) + αβ = 16 - 4(3) + 1 = 16 - 12 + 1 = 5. S' = 25 / 5 = 5. 6. Product of new roots (P') = R&sub1; * R&sub2; = [(3α - 1) / (4 - α)] * [(3β - 1) / (4 - β)] P' = (9αβ - 3α - 3β + 1) / (16 - 4α - 4β + αβ) P' = [9αβ - 3(α + β) + 1] / [16 - 4(α + β) + αβ] Substitute values: P' = [9(1) - 3(3) + 1] / [16 - 4(3) + 1] = [9 - 9 + 1] / [16 - 12 + 1] = 1 / 5. 7. The new quadratic equation is x² - S'x + P' = 0. x² - 5x + 1/5 = 0. Multiplying by 5: 5x² - 25x + 1 = 0.
Final Answer: 5x² - 25x + 1 = 0
Problem 255
Hard 4 Marks
If α and β are the roots of the quadratic equation x² - 2x + 4 = 0, form a quadratic equation whose roots are α³ and β³.
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1. For the given equation x&sup2; - 2x + 4 = 0, we have: Sum of roots (&alpha; + &beta;) = -(-2)/1 = 2. Product of roots (&alpha;&beta;) = 4/1 = 4. 2. We need to find the sum and product of the new roots, which are &alpha;&sup3; and &beta;&sup3;. Sum of new roots (S') = &alpha;&sup3; + &beta;&sup3;. Product of new roots (P') = &alpha;&sup3;&beta;&sup3; = (&alpha;&beta;)&sup3;. 3. Calculate S': &alpha;&sup3; + &beta;&sup3; = (&alpha; + &beta;)(&alpha;&sup2; - &alpha;&beta; + &beta;&sup2;) &alpha;&sup3; + &beta;&sup3; = (&alpha; + &beta;)[(&alpha; + &beta;)&sup2; - 3&alpha;&beta;] Substitute the values of (&alpha; + &beta;) and &alpha;&beta;: S' = (2)[(2)&sup2; - 3(4)] = 2[4 - 12] = 2[-8] = -16. 4. Calculate P': P' = (&alpha;&beta;)&sup3; = (4)&sup3; = 64. 5. The new quadratic equation is given by x&sup2; - S'x + P' = 0. Substituting S' and P': x&sup2; - (-16)x + 64 = 0. x&sup2; + 16x + 64 = 0. <br><b>Alternative approach:</b> 1. Multiply the given equation x&sup2; - 2x + 4 = 0 by (x+2): (x+2)(x&sup2; - 2x + 4) = 0 x&sup3; + 8 = 0 So, x&sup3; = -8. 2. Since &alpha; and &beta; are roots of x&sup2; - 2x + 4 = 0, they must also satisfy x&sup3; = -8. Therefore, &alpha;&sup3; = -8 and &beta;&sup3; = -8. 3. The new roots are -8 and -8. 4. Sum of new roots (S') = -8 + (-8) = -16. 5. Product of new roots (P') = (-8)(-8) = 64. 6. The new quadratic equation is x&sup2; - S'x + P' = 0. x&sup2; - (-16)x + 64 = 0. x&sup2; + 16x + 64 = 0.
Final Answer: x&sup2; + 16x + 64 = 0
Problem 255
Medium 4 Marks
Form the quadratic equation whose roots are tan(15°) and cot(15°).
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1. Evaluate tan(15°): tan(15°) = tan(45° - 30°). 2. Use the tangent subtraction formula: tan(A - B) = (tanA - tanB) / (1 + tanA tanB). 3. tan(15°) = (tan45° - tan30°) / (1 + tan45° tan30°) = (1 - 1/√3) / (1 + 1*1/√3) = (√3 - 1)/(√3 + 1). 4. Rationalize the denominator: tan(15°) = ((√3 - 1)(√3 - 1)) / ((√3 + 1)(√3 - 1)) = (3 + 1 - 2√3) / (3 - 1) = (4 - 2√3) / 2 = 2 - √3. 5. Evaluate cot(15°): cot(15°) = 1/tan(15°). 6. cot(15°) = 1/(2 - √3). Rationalize: (1 * (2 + √3)) / ((2 - √3)(2 + √3)) = (2 + √3) / (4 - 3) = 2 + √3. 7. Let the roots be r₁ = 2 - √3 and r₂ = 2 + √3. 8. Calculate the sum of the roots: S = r₁ + r₂ = (2 - √3) + (2 + √3) = 4. 9. Calculate the product of the roots: P = r₁ * r₂ = (2 - √3)(2 + √3) = 2² - (√3)² = 4 - 3 = 1. 10. The required quadratic equation is x² - Sx + P = 0. 11. Substitute S = 4 and P = 1: x² - 4x + 1 = 0.
Final Answer: x² - 4x + 1 = 0
Problem 255
Medium 4 Marks
If α and β are the roots of the equation x² - 6x + 2 = 0, form the quadratic equation whose roots are (α/β + β/α) and (α/β)² + (β/α)².
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1. From the given equation, α + β = 6 and αβ = 2. 2. Let the new roots be y₁ = α/β + β/α and y₂ = (α/β)² + (β/α)². 3. Calculate y₁: y₁ = (α² + β²)/(αβ) = ((α + β)² - 2αβ)/(αβ). 4. Substitute α + β = 6 and αβ = 2: y₁ = (6² - 2*2)/2 = (36 - 4)/2 = 32/2 = 16. 5. Calculate y₂: y₂ = (α/β)² + (β/α)². This can be written as (α/β + β/α)² - 2, which is y₁² - 2. 6. Substitute y₁ = 16: y₂ = 16² - 2 = 256 - 2 = 254. 7. Calculate the sum of the new roots: S' = y₁ + y₂ = 16 + 254 = 270. 8. Calculate the product of the new roots: P' = y₁ * y₂ = 16 * 254. 9. P' = 4064. 10. The required quadratic equation is x² - S'x + P' = 0. 11. Substitute S' = 270 and P' = 4064: x² - 270x + 4064 = 0.
Final Answer: x² - 270x + 4064 = 0
Problem 255
Easy 4 Marks
Form the quadratic equation whose roots are 5 and -2.
Show Solution
1. Calculate the sum of the roots: S = α + β = 5 + (-2) = 3. 2. Calculate the product of the roots: P = αβ = 5 × (-2) = -10. 3. Use the general form of a quadratic equation: x² - Sx + P = 0. 4. Substitute the values of S and P: x² - (3)x + (-10) = 0. 5. Simplify the equation: x² - 3x - 10 = 0.
Final Answer: x² - 3x - 10 = 0
Problem 255
Medium 4 Marks
If one root of the quadratic equation ax² + bx + c = 0 is the square of the other, then prove that b³ + a²c + ac² - 3abc = 0.
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1. Let the roots of ax² + bx + c = 0 be r and r². 2. From Vieta's formulas: Sum of roots: r + r² = -b/a. Product of roots: r * r² = r³ = c/a. 3. From r³ = c/a, we have r = (c/a)^(1/3). 4. Substitute the value of r into the sum of roots equation: (c/a)^(1/3) + (c/a)^(2/3) = -b/a. 5. Multiply the entire equation by 'a' to clear denominators and simplify powers: a * (c/a)^(1/3) + a * (c/a)^(2/3) = -b. 6. This simplifies to a^(2/3)c^(1/3) + a^(1/3)c^(2/3) = -b. 7. Cube both sides of the equation: [a^(2/3)c^(1/3) + a^(1/3)c^(2/3)]³ = (-b)³. 8. Use the algebraic identity (X+Y)³ = X³ + Y³ + 3XY(X+Y). Here, X = a^(2/3)c^(1/3) and Y = a^(1/3)c^(2/3). 9. X³ = (a^(2/3)c^(1/3))³ = a²c. Y³ = (a^(1/3)c^(2/3))³ = ac². 10. XY = (a^(2/3)c^(1/3))(a^(1/3)c^(2/3)) = a^(2/3+1/3)c^(1/3+2/3) = ac. 11. Substitute X, Y, XY, and (X+Y = -b) into the cubic identity: a²c + ac² + 3(ac)(-b) = -b³. 12. Simplify: a²c + ac² - 3abc = -b³. 13. Rearrange to get the desired relation: b³ + a²c + ac² - 3abc = 0.
Final Answer: b³ + a²c + ac² - 3abc = 0
Problem 255
Medium 4 Marks
If α and β are the roots of the equation x² - 3x + 1 = 0, form the quadratic equation whose roots are 1/(α-2) and 1/(β-2).
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1. From the given equation, α + β = 3 and αβ = 1. 2. Let the new roots be y₁ = 1/(α-2) and y₂ = 1/(β-2). 3. Calculate the sum of the new roots: S' = y₁ + y₂ = 1/(α-2) + 1/(β-2) = (β-2 + α-2) / ((α-2)(β-2)). 4. Simplify S': S' = (α+β-4) / (αβ - 2(α+β) + 4). 5. Substitute α+β=3 and αβ=1: S' = (3-4) / (1 - 2(3) + 4) = -1 / (1 - 6 + 4) = -1 / -1 = 1. 6. Calculate the product of the new roots: P' = y₁ * y₂ = 1/((α-2)(β-2)). 7. Simplify P': P' = 1 / (αβ - 2(α+β) + 4). 8. Substitute α+β=3 and αβ=1: P' = 1 / (1 - 2(3) + 4) = 1 / (1 - 6 + 4) = 1 / -1 = -1. 9. The required quadratic equation is x² - S'x + P' = 0. 10. Substitute S' and P': x² - (1)x + (-1) = 0, which simplifies to x² - x - 1 = 0.
Final Answer: x² - x - 1 = 0
Problem 255
Medium 4 Marks
If α and β are the roots of the equation x² - px + q = 0, form the quadratic equation whose roots are α²/β and β²/α.
Show Solution
1. From the given equation, α + β = p and αβ = q. 2. Let the new roots be y₁ = α²/β and y₂ = β²/α. 3. Calculate the sum of the new roots: S' = y₁ + y₂ = α²/β + β²/α = (α³ + β³)/(αβ). 4. Use the identity α³ + β³ = (α + β)((α + β)² - 3αβ). 5. Substitute the values of α + β and αβ: S' = [p(p² - 3q)]/q. 6. Calculate the product of the new roots: P' = y₁ * y₂ = (α²/β) * (β²/α) = αβ. 7. Substitute the value of αβ: P' = q. 8. The required quadratic equation is x² - S'x + P' = 0. 9. Substitute S' and P': x² - [p(p² - 3q)/q]x + q = 0. 10. Multiply by q to clear the denominator: qx² - p(p² - 3q)x + q² = 0.
Final Answer: qx² - p(p² - 3q)x + q² = 0
Problem 255
Easy 4 Marks
A quadratic equation has roots such that their sum is 6 and their product is 9. Find the equation.
Show Solution
1. The problem directly provides the sum (S) and product (P) of the roots. 2. Use the general form of a quadratic equation: x² - Sx + P = 0. 3. Substitute the given values of S and P: x² - (6)x + (9) = 0. 4. Simplify the equation: x² - 6x + 9 = 0.
Final Answer: x² - 6x + 9 = 0
Problem 255
Easy 4 Marks
If the roots of the equation x² - 7x + 10 = 0 are α and β, find the quadratic equation whose roots are α+1 and β+1.
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1. From the given equation x² - 7x + 10 = 0, identify the sum and product of roots: α + β = -(-7)/1 = 7 and αβ = 10/1 = 10. 2. For the new roots (α+1) and (β+1): a. Calculate the new sum of roots (S'): S' = (α+1) + (β+1) = α + β + 2 = 7 + 2 = 9. b. Calculate the new product of roots (P'): P' = (α+1)(β+1) = αβ + α + β + 1 = 10 + 7 + 1 = 18. 3. Use the general form for the new equation: x² - S'x + P' = 0. 4. Substitute S' and P': x² - (9)x + (18) = 0.
Final Answer: x² - 9x + 18 = 0
Problem 255
Easy 4 Marks
If 2 + 3i is a root of a quadratic equation with real coefficients, find the equation.
Show Solution
1. Since the coefficients are real and one root is 2 + 3i, the other root (β) must be its complex conjugate: β = 2 - 3i. 2. Calculate the sum of the roots: S = (2 + 3i) + (2 - 3i) = 2 + 2 + 3i - 3i = 4. 3. Calculate the product of the roots: P = (2 + 3i)(2 - 3i) = 2² - (3i)² = 4 - 9i² = 4 - 9(-1) = 4 + 9 = 13. 4. Use the general form: x² - Sx + P = 0. 5. Substitute S and P: x² - (4)x + (13) = 0. 6. Simplify: x² - 4x + 13 = 0.
Final Answer: x² - 4x + 13 = 0
Problem 255
Easy 4 Marks
Form a quadratic equation whose roots are 1/2 and -3/4.
Show Solution
1. Calculate the sum of the roots: S = 1/2 + (-3/4) = 2/4 - 3/4 = -1/4. 2. Calculate the product of the roots: P = (1/2) × (-3/4) = -3/8. 3. Use the general form: x² - Sx + P = 0. 4. Substitute S and P: x² - (-1/4)x + (-3/8) = 0. 5. Simplify and clear denominators: x² + (1/4)x - 3/8 = 0. Multiply by 8 to get 8x² + 2x - 3 = 0.
Final Answer: 8x² + 2x - 3 = 0
Problem 255
Easy 4 Marks
If the roots of a quadratic equation are (1 + √2) and (1 - √2), find the equation.
Show Solution
1. Calculate the sum of the roots: S = (1 + √2) + (1 - √2) = 1 + 1 + √2 - √2 = 2. 2. Calculate the product of the roots: P = (1 + √2)(1 - √2) = 1² - (√2)² = 1 - 2 = -1. 3. Use the general form: x² - Sx + P = 0. 4. Substitute S and P: x² - (2)x + (-1) = 0. 5. Simplify: x² - 2x - 1 = 0.
Final Answer: x² - 2x - 1 = 0

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📐Important Formulas (3)

Formation of Quadratic Equation from Roots
x^2 - (alpha + eta)x + alphaeta = 0
Text: x^2 - (Sum of Roots)x + (Product of Roots) = 0
This is the fundamental formula to construct a quadratic equation when its roots are given. If (alpha) and (eta) are the two roots of a quadratic equation, then the equation can be expressed in the form (x^2 - Sx + P = 0), where (S = alpha + eta) is the sum of the roots and (P = alphaeta) is the product of the roots. This formula directly links the roots of an equation to its coefficients.
Variables: Use this formula whenever you are given the values of the roots ((alpha) and (eta)) and are asked to find the corresponding quadratic equation. It is also useful when you know the sum and product of the roots directly.
Sum of Roots
S = alpha + eta
Text: Sum of Roots = (alpha + eta)
The sum of the roots, denoted by (S), is simply the addition of the two given roots (alpha) and (eta). In a general quadratic equation (ax^2 + bx + c = 0), the sum of the roots is also given by (-b/a). This identity is crucial for relating roots to coefficients.
Variables: Calculate the sum of roots as the first step when forming a quadratic equation from given roots. Also used to find the sum of roots of an existing quadratic equation without solving for individual roots.
Product of Roots
P = alphaeta
Text: Product of Roots = (alpha imes eta)
The product of the roots, denoted by (P), is the multiplication of the two given roots (alpha) and (eta). For a general quadratic equation (ax^2 + bx + c = 0), the product of the roots is also given by (c/a). This provides another vital link between the roots and the coefficients.
Variables: Calculate the product of roots as the second step when forming a quadratic equation from given roots. Also used to find the product of roots of an existing quadratic equation without solving for individual roots.

📚References & Further Reading (10)

Book
Algebra for JEE Main & Advanced
By: G. Tewani
Not applicable for physical book, check Cengage India website for product details.
A comprehensive textbook designed for JEE Main and Advanced aspirants. It covers the topic of quadratic equations in detail, including advanced techniques for forming equations from given roots, with numerous solved examples and practice problems relevant to competitive exams.
Note: Highly relevant for JEE preparation, covers advanced aspects and problem-solving strategies essential for competitive exams, building on foundational knowledge.
Book
By:
Website
Forming Quadratic Equations
By: MathsIsFun.com
https://www.mathsisfun.com/algebra/quadratic-equation-from-roots.html
A straightforward and concise explanation of how to form quadratic equations when the roots are given. It uses simple language, illustrative examples, and often includes interactive elements or self-check questions.
Note: Good for quick review and reinforcement of the topic. The clear presentation and examples aid in solidifying understanding.
Website
By:
PDF
Algebra Review: Quadratic Equations and Functions
By: University of Houston, Department of Mathematics
https://online.math.uh.edu/Math1300/PDF/Algebra_QuadraticEquations.pdf
Part of a comprehensive algebra review, this PDF section clearly outlines the relationship between roots and coefficients, and the method to construct quadratic equations given their roots. It includes explanations and simple examples.
Note: Reliable academic resource for a solid theoretical understanding. Good for students seeking a more formal, yet accessible, explanation of the concepts.
PDF
By:
Article
Quadratic Equations: Introduction, Methods of Solving, Roots and Relationship
By: Vedantu Learning Centre
https://www.vedantu.com/maths/quadratic-equations
An online article providing a detailed overview of quadratic equations, including a dedicated section on forming quadratic equations when the roots are known. It often includes examples tailored for Indian curriculum students.
Note: Covers the topic comprehensively within a broader context of quadratic equations. Useful for quick reference and reinforcing concepts taught in class.
Article
By:
Research_Paper
The History of Quadratic Equations
By: Katz, Victor J.
https://math.stackexchange.com/questions/285521/references-on-the-history-of-quadratic-equations (Reference to a discussion where such historical papers are cited)
A research paper or academic essay exploring the historical development of quadratic equations, from ancient civilizations to modern algebra. It may implicitly cover how mathematicians historically understood and formulated equations based on their solutions or conditions.
Note: Provides a broader historical context, enriching understanding of the topic's evolution. While not directly instructional, it offers a deeper appreciation for the mathematical concepts.
Research_Paper
By:

⚠️Common Mistakes to Avoid (58)

Minor Other

❌ Incorrect Sign for Sum of Roots Term

Students frequently make a sign error when forming a quadratic equation using the sum and product of its roots. Instead of using the negative of the sum of roots as the coefficient of 'x', they often incorrectly use the positive sum.
💭 Why This Happens:
This mistake typically arises from rote memorization of the formula without understanding its derivation, or simple algebraic oversight under exam pressure. The standard form ax2 + bx + c = 0 can be written as x2 - (-b/a)x + (c/a) = 0. Here, (-b/a) is the sum of roots, and (c/a) is the product. Students often forget the inherent negative sign associated with the sum term in the simplified x2 - (Sum)x + (Product) = 0 form.
✅ Correct Approach:
Always remember the standard form of a quadratic equation whose roots α and β are given: x2 - (α + β)x + (αβ) = 0. The coefficient of 'x' is always the negative of the sum of the roots.
📝 Examples:
❌ Wrong:
If the roots are α = 2 and β = 3:
Sum of roots = 2 + 3 = 5
Product of roots = 2 * 3 = 6
Incorrect Equation: x2 + 5x + 6 = 0
✅ Correct:
If the roots are α = 2 and β = 3:
Sum of roots = 2 + 3 = 5
Product of roots = 2 * 3 = 6
Correct Equation: x2 - 5x + 6 = 0 (which is (x-2)(x-3)=0)
💡 Prevention Tips:
  • Recall the Derivation: Remember that if α and β are roots, then (x - α)(x - β) = 0 is the equation. Expanding this gives x2 - (α + β)x + αβ = 0, clearly showing the negative sign.
  • Mindful Substitution: Before substituting, explicitly write down x2 - (Sum of Roots)x + (Product of Roots) = 0.
  • Practice with Varied Roots: Work through problems involving both positive and negative roots to solidify understanding of sign conventions.
JEE_Advanced
Minor Conceptual

❌ Incorrect Signs or Placement of Sum/Product of Roots

Students frequently make sign errors, particularly with the coefficient of the 'x' term (which is related to the sum of roots), or they might incorrectly interchange the positions of the sum and product of roots in the general quadratic equation formula. Common incorrect forms include $x^2 + (alpha+eta)x + alphaeta = 0$ or $x^2 - alphaeta x + (alpha+eta) = 0$.
💭 Why This Happens:
This mistake primarily arises from rote memorization of the formula without a clear understanding of its derivation from $(x-alpha)(x-eta)=0$. The standard quadratic equation $ax^2+bx+c=0$ has the sum of roots as $-b/a$ and product as $c/a$. When we normalize it to a monic quadratic ($a=1$), it becomes $x^2 - ( ext{sum})x + ( ext{product}) = 0$. The negative sign before the sum of roots is a critical detail often overlooked.
✅ Correct Approach:
Always refer to the standard and correct form for a quadratic equation with roots $alpha$ and $eta$: $x^2 - (alpha + eta)x + alphaeta = 0$.

  • Step 1: Calculate the sum of the roots ($alpha + eta$).

  • Step 2: Calculate the product of the roots ($alpha eta$).

  • Step 3: Substitute these values precisely into the formula, ensuring the negative sign for the sum of roots and the positive sign for the product of roots.

📝 Examples:
❌ Wrong:
Given roots are $2$ and $-3$.

Wrong Approach:



  • Sum of roots: $2 + (-3) = -1$

  • Product of roots: $2 imes (-3) = -6$

  • Incorrect equation formed: $x^2 + ( ext{Sum})x + ( ext{Product}) = 0$ (Mistake in the sign for the middle term)

  • $Rightarrow x^2 + (-1)x + (-6) = 0$

  • $x^2 - x - 6 = 0$


This is incorrect as the middle term's sign is opposite to the correct one.

✅ Correct:
Given roots are $2$ and $-3$.

Correct Approach:



  • Sum of roots ($alpha + eta$): $2 + (-3) = -1$

  • Product of roots ($alpha eta$): $2 imes (-3) = -6$

  • Using the formula $x^2 - (alpha + eta)x + alphaeta = 0$:

  • $x^2 - (-1)x + (-6) = 0$

  • $x^2 + x - 6 = 0$


This is the correct quadratic equation.

💡 Prevention Tips:

  • Understand the Derivation: Always remember that the formula is derived from $(x-alpha)(x-eta)=0$, which expands to $x^2 - (alpha+eta)x + alphaeta = 0$. This clarifies why the sum term has a negative coefficient.

  • Double-Check Signs: Make it a habit to double-check the signs for both the sum and product terms immediately after substitution.

  • Practice Regularly: Solve a diverse range of problems, especially those involving negative or complex roots, to ingrain the correct application of the formula.

  • JEE Main Tip: While seemingly minor, these conceptual sign errors are common in JEE Main under time pressure. A quick mental or written check after forming the equation can prevent losing easy marks.

JEE_Main
Minor Formula

❌ Incorrect Sign for the Sum of Roots

A common error is to use an incorrect sign for the term involving the sum of roots when forming a quadratic equation. Instead of the required negative sign, students sometimes use a positive sign, leading to an incorrect quadratic equation.
💭 Why This Happens:
This mistake often arises from misremembering the standard formula for forming a quadratic equation. The general form x² - (Sum of Roots)x + (Product of Roots) = 0 has a crucial negative sign before the sum of roots. Students might mistakenly recall it as x² + (Sum of Roots)x + (Product of Roots) = 0, or simply overlook the negative sign due to haste.
✅ Correct Approach:
Always ensure to use the correct standard form of a quadratic equation when given its roots (let's say α and β):
x² - (α + β)x + (αβ) = 0.
Here, α + β represents the sum of the roots, and αβ represents the product of the roots. The coefficient of the 'x' term is always the negative of the sum of the roots.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3:
Sum of roots (α + β) = 2 + 3 = 5
Product of roots (αβ) = 2 × 3 = 6
Incorrect equation: x² + (5)x + 6 = 0 → x² + 5x + 6 = 0.
✅ Correct:
Using the same roots, 2 and 3:
Sum of roots (α + β) = 5
Product of roots (αβ) = 6
Correct equation: x² - (5)x + 6 = 0 → x² - 5x + 6 = 0.
(CBSE & JEE: This basic formula is fundamental and frequently tested directly or indirectly).
💡 Prevention Tips:
  • Memorize the Formula Precisely: Commit the formula x² - (Sum of Roots)x + (Product of Roots) = 0 to memory, emphasizing the negative sign for the sum of roots.
  • Verify with a Quick Check: After forming the equation, mentally substitute one of the given roots back into it. If the equation evaluates to zero, your formula application is likely correct.
  • Practice Regularly: Consistent practice with various examples helps in solidifying the correct formula recall.
JEE_Main
Minor Unit Conversion

❌ <span style='color: #FF0000;'>Misapplication of "Conversion" from Roots to Coefficients (Minor Algebraic Calculation Error)</span>

Students sometimes make minor algebraic errors when 'converting' (calculating) the sum (α + β) or product (αβ) of the given roots into the coefficients of the quadratic equation. While true unit conversion (e.g., meters to centimeters) is not typically relevant in forming quadratic equations from abstract numerical roots, students might conceptually falter in 'converting' the *form* of the roots (e.g., negative, fractional, or radical values) into their correctly calculated sum and product. These are essentially careless arithmetic and sign errors, which are minor but lead to an incorrect final equation.
💭 Why This Happens:
  1. Careless Arithmetic: Simple addition/subtraction or multiplication errors, especially with fractions or negative numbers.
  2. Sign Errors: Incorrectly handling negative signs when computing S = α + β or P = αβ.
  3. Formula Misapplication: Occasionally, mixing up the signs in the standard formula x² - Sx + P = 0, for example, using +Sx instead of -Sx.
✅ Correct Approach:
The fundamental step is to correctly calculate the sum of roots (S = α + β) and the product of roots (P = αβ) with utmost care. After obtaining these values, substitute them into the standard form of the quadratic equation: x² - Sx + P = 0. Always double-check calculations, particularly when dealing with negative numbers, fractions, or radicals.

CBSE vs. JEE: The core principle is identical. JEE questions might involve more complex roots (e.g., surds, complex numbers) requiring more rigorous algebraic manipulation, increasing the chances of minor calculation errors.
📝 Examples:
❌ Wrong:
Given roots α = 1/2 and β = -3.
Incorrect calculation of sum: S = 1/2 + (-3) = 1/2 - 3 = -2.5 (instead of -2.5, a student might incorrectly calculate it as +2.5 or other wrong value due to careless sign/fraction handling).
Product P = (1/2) * (-3) = -3/2.
Resulting incorrect equation (if sum was mistakenly +2.5): x² - 2.5x - 1.5 = 0.
✅ Correct:
Given roots α = 1/2 and β = -3.
Correct calculation:
Sum S = α + β = 1/2 + (-3) = 1/2 - 3 = 1/2 - 6/2 = -5/2.
Product P = αβ = (1/2) * (-3) = -3/2.
Using the formula x² - Sx + P = 0:
x² - (-5/2)x + (-3/2) = 0
x² + (5/2)x - (3/2) = 0
Multiplying by 2 to clear denominators: 2x² + 5x - 3 = 0.
💡 Prevention Tips:
  • Systematic Calculation: Always write down S = α + β and P = αβ clearly and calculate them step-by-step before substitution.
  • Vigilance with Signs: Pay extra attention to negative signs, especially when adding or multiplying roots. A common error is -(-S) becoming -S.
  • Fraction and Radical Mastery: Ensure strong command over arithmetic operations involving fractions, decimals, and radicals to avoid trivial mistakes.
  • Formula Check: Double-check the formula x² - Sx + P = 0 itself, ensuring the sign of the 'Sx' term is correct.
  • Self-Verification: If time permits, find the roots of your derived quadratic equation to confirm they match the given initial roots. This is a powerful self-check.
JEE_Main
Minor Sign Error

❌ Sign Error in the Standard Form (x² - Sx + P = 0)

A common minor mistake is incorrectly applying the sign of the sum of roots when forming a quadratic equation. Students often write `x² + (sum of roots)x + (product of roots) = 0` instead of the correct `x² - (sum of roots)x + (product of roots) = 0`.
💭 Why This Happens:
This error primarily stems from haste, a lack of clear understanding of the formula's derivation, or confusion with Vieta's formulas where the sum of roots for `ax² + bx + c = 0` is `-b/a`. The standard form `x² - (α+β)x + αβ = 0` directly comes from expanding `(x-α)(x-β) = 0`, where the negative sign naturally appears with the sum of roots term.
✅ Correct Approach:
The correct and universally accepted standard form of a quadratic equation with roots α and β is given by:
x² - (α + β)x + (αβ) = 0
Alternatively, if S is the sum of roots and P is the product of roots, the equation is:
x² - Sx + P = 0
Always remember the negative sign before the sum of roots term.
📝 Examples:
❌ Wrong:
If the roots are 3 and -5:
A student might calculate Sum (S) = 3 + (-5) = -2, Product (P) = 3 * (-5) = -15.
Then incorrectly write the equation as: x² + (-2)x + (-15) = 0 which simplifies to x² - 2x - 15 = 0.
✅ Correct:
Using the same roots, 3 and -5:
Sum (S) = 3 + (-5) = -2
Product (P) = 3 * (-5) = -15
Applying the correct formula x² - Sx + P = 0:
x² - (-2)x + (-15) = 0
This correctly simplifies to: x² + 2x - 15 = 0.
💡 Prevention Tips:
  • Memorize the Formula Precisely: Commit x² - (sum of roots)x + (product of roots) = 0 to memory.
  • Calculate Separately: Always calculate the sum of roots (S) and product of roots (P) first, then substitute these values into the formula.
  • Double Check: Before finalizing the answer, explicitly check the sign of the 'x' term.
  • Understand Derivation: Remember that the formula comes from `(x-α)(x-β) = 0`, which directly leads to `x² - (α+β)x + αβ = 0`.
JEE_Main
Minor Approximation

❌ Premature Approximation of Irrational Numbers in Roots

Students sometimes approximate the values of irrational numbers (like √2, √3, etc.) present within the given roots *before* calculating the sum and product of roots. This leads to slightly inaccurate values for the sum and product, resulting in a quadratic equation with coefficients that are not exactly what's expected for JEE Main, where exact answers are typically required.
💭 Why This Happens:
  • A misguided attempt to simplify arithmetic, especially when dealing with complex expressions involving square roots.
  • Overlooking the precision required for JEE Main problems, where exact irrational forms should be maintained until the final step, or if specifically asked for an approximate answer.
  • Confusion between board exam problems (where approximations might be allowed for final answers) and JEE Main (which demands exactness unless otherwise specified).
✅ Correct Approach:
  • Always work with the exact forms of irrational numbers within the roots.
  • Calculate the sum (S = α + β) and product (P = αβ) using these exact forms.
  • Substitute the exact S and P into the standard quadratic equation formula: x² - Sx + P = 0.
  • Only approximate if the problem explicitly asks for an approximate answer or if the final context demands it (very rare for equation formation in JEE).
📝 Examples:
❌ Wrong:

Given roots: α = 1 + √2, β = 3

Student might approximate √2 ≈ 1.414

  • Approximated α: α ≈ 1 + 1.414 = 2.414
  • Sum of roots (S): S ≈ 2.414 + 3 = 5.414
  • Product of roots (P): P ≈ 2.414 × 3 = 7.242
  • Resulting approximate equation: x² - 5.414x + 7.242 = 0 (Incorrect for JEE Main)
✅ Correct:

Given roots: α = 1 + √2, β = 3

  • Sum of roots (S): S = (1 + √2) + 3 = 4 + √2
  • Product of roots (P): P = (1 + √2) × 3 = 3 + 3√2
  • Correct quadratic equation: x² - (4 + √2)x + (3 + 3√2) = 0

JEE Main specific: For multiple-choice questions, the options will typically contain the exact irrational forms, making approximation a dead-end.

💡 Prevention Tips:
  • Stay Exact: Always use the precise irrational forms (e.g., √2, √3) until the very last step.
  • Practice Algebraic Manipulation: Familiarize yourself with addition, subtraction, multiplication, and division of irrational numbers to avoid needing to approximate.
  • Check Options Carefully: In MCQ format, if options contain exact irrational values, it's a strong hint that approximation is not the intended path.
  • CBSE vs. JEE Mindset: Understand that while CBSE might sometimes allow approximations for final answers in certain contexts, JEE Main almost always demands exact values unless explicitly specified.
JEE_Main
Minor Other

❌ Omitting the General Multiplicative Constant 'k'

Students often correctly apply the formula x2 - (Sum of Roots)x + (Product of Roots) = 0 to form a quadratic equation from given roots. However, they frequently forget that any non-zero constant 'k' can multiply this entire equation, i.e., k[x2 - (Sum of Roots)x + (Product of Roots)] = 0, without changing its roots. This omission can be a minor conceptual oversight but crucial for questions that require the most general form or when additional conditions (like passing through a specific point) are provided to determine 'k'.
💭 Why This Happens:
This mistake stems from an over-reliance on the standard monic form (where the leading coefficient 'a' is implicitly 1) and not fully grasping that multiplying a polynomial by a constant does not alter its roots. Many students memorize the basic formula without understanding its derivation or its general form.
✅ Correct Approach:
Always consider the general form of a quadratic equation with roots α and β as k[x2 - (α+β)x + αβ] = 0 or k(x - α)(x - β) = 0, where k ≠ 0. If no additional information is provided, setting k=1 gives the simplest quadratic equation. However, in JEE Main, if a condition is given (e.g., the parabola passes through a specific point, or its leading coefficient is specified), 'k' must be determined from that condition.
📝 Examples:
❌ Wrong:
Given roots 2 and 3, a student incorrectly states:
  • The only quadratic equation is x2 - (2+3)x + (2×3) = 0, which simplifies to x2 - 5x + 6 = 0.
This is a quadratic equation, but not necessarily the only one or the most general form.
✅ Correct:
Given roots 2 and 3:
  • The general form of a quadratic equation with these roots is k(x2 - 5x + 6) = 0, where k ≠ 0.
  • JEE Application: If the question further states that the parabola passes through the point (1, 4), then substitute these values into the general equation:k((1)2 - 5(1) + 6) = 4k(1 - 5 + 6) = 4k(2) = 4 ⇒ k = 2Thus, the specific quadratic equation is 2(x2 - 5x + 6) = 0 or 2x2 - 10x + 12 = 0.
💡 Prevention Tips:
  • Understand the General Form: Always remember that k is a non-zero constant multiplying the entire quadratic expression.
  • Read Carefully: Pay close attention to wording like 'a quadratic equation' (simplest form usually suffices) versus 'the quadratic equation' (implies unique 'k' needs to be found) or if additional points are given.
  • Include 'k' as Placeholder: Initially include 'k' in your setup, even if you anticipate it might be 1. This reinforces the concept and prevents errors when specific conditions are given.
JEE_Main
Minor Other

❌ <span style='color: #FF0000;'>Leaving coefficients as fractions or unsimplified.</span>

Students often correctly find the sum and product of the given roots and substitute them into the formula x² - (Sum of Roots)x + (Product of Roots) = 0. However, they sometimes leave the quadratic equation with fractional coefficients instead of converting it into an equivalent equation with integer coefficients, which is the standard and preferred form for CBSE examinations.
💭 Why This Happens:
This typically occurs due to an oversight or a lack of emphasis on the standard form of a quadratic equation (Ax² + Bx + C = 0, where A, B, C are integers and A ≠ 0). Students might consider `x² - (5/6)x + (1/6) = 0` as a complete answer, not realizing it can be further simplified. This is a minor error in presentation and standard practice rather than a conceptual misunderstanding.
✅ Correct Approach:
After substituting the sum and product of roots into the formula, multiply the entire equation by the least common multiple (LCM) of the denominators of the fractional coefficients to clear all fractions and obtain an equation with integer coefficients. Ensure the leading coefficient is usually positive, or at least the equation is in its simplest integer form. For CBSE, presenting the simplest integer coefficient form is generally expected.
📝 Examples:
❌ Wrong: 
Given roots: 1/2 and 1/3.
Sum of roots = 1/2 + 1/3 = 5/6
Product of roots = (1/2) * (1/3) = 1/6
Incorrect Quadratic equation: x² - (5/6)x + (1/6) = 0
✅ Correct: 
Given roots: 1/2 and 1/3.
Sum of roots = 1/2 + 1/3 = 5/6
Product of roots = (1/2) * (1/3) = 1/6
Quadratic equation: x² - (5/6)x + (1/6) = 0

Step to correct: Multiply the entire equation by the LCM of denominators (6):
6 * (x² - (5/6)x + (1/6)) = 6 * 0
Correct Quadratic equation: 6x² - 5x + 1 = 0
💡 Prevention Tips:
  • Always check the final form: After forming the equation, quickly review if all coefficients are integers.
  • Clear fractions: If fractional coefficients are present, multiply the entire equation by the LCM of their denominators to eliminate fractions.
  • Simplify: Ensure there are no common factors among the coefficients that can be divided out (e.g., `2x² - 4x + 2 = 0` should be simplified to `x² - 2x + 1 = 0`).
  • Standard practice: Always aim for the simplest integer coefficients unless the question specifies otherwise.
CBSE_12th
Minor Approximation

❌ Premature Approximation of Irrational/Non-Integer Roots

Students sometimes approximate irrational roots (e.g., √2 ≈ 1.414) or non-integer roots (e.g., 2/3 ≈ 0.667) prematurely before calculating their sum and product. This leads to a quadratic equation with approximate decimal coefficients instead of exact, often rational or surd, coefficients.
💭 Why This Happens:
This error stems from a misunderstanding that all numbers must be converted into decimal form for calculations. Students might feel it's 'neater' or easier to work with decimals, or they may struggle with the algebraic manipulation of surds or fractions. They overlook the instruction to find the 'exact' quadratic equation.
✅ Correct Approach:
Always perform operations (sum and product of roots) using the exact forms of the roots, whether they are irrational (surds) or fractions. Coefficients of a quadratic equation formed from given roots are expected to be exact, usually rational or in simplest surd form. Only approximate if the question explicitly asks for an approximate equation or approximate roots in the final answer.
📝 Examples:
❌ Wrong:
Given roots are 1 and 1 + √2.
A student might approximate √2 ≈ 1.414.
Sum (S) ≈ 1 + (1 + 1.414) = 1 + 2.414 = 3.414
Product (P) ≈ 1 * (1 + 1.414) = 1 * 2.414 = 2.414
Approximate Equation: x² - 3.414x + 2.414 = 0. This is an incorrect form for the exact equation.
✅ Correct:
Given roots are 1 and 1 + √2.
Sum (S) = 1 + (1 + √2) = 2 + √2
Product (P) = 1 * (1 + √2) = 1 + √2
Correct Equation: x² - (2 + √2)x + (1 + √2) = 0. (For CBSE, coefficients are often rationalized or simplified further if possible, but this is the exact form).
💡 Prevention Tips:
  • Work with Exact Values: Always keep irrational numbers in their surd form (e.g., √3) and fractions as fractions (e.g., 5/7) throughout the calculation of sum and product.
  • Read the Question Carefully: Look for keywords like 'exact equation' or 'approximate to N decimal places'. If not specified, assume exact form.
  • Practice Surd and Fraction Arithmetic: Strengthen your skills in adding, subtracting, multiplying, and dividing expressions involving surds and fractions to avoid the temptation of using decimals.
CBSE_12th
Minor Sign Error

❌ Sign Errors in Determining Coefficients

Students frequently make sign errors when forming a quadratic equation using the standard formula x² - (Sum of Roots)x + (Product of Roots) = 0. The most common mistake involves mismanaging the negative sign before the 'Sum of Roots' term, especially when the sum itself is negative. This leads to an incorrect sign for the coefficient of 'x'.
💭 Why This Happens:
This error primarily stems from a lack of careful attention to the general form of the quadratic equation. Students often forget the inherent negative sign in the formula (- (α + β)x) or perform incorrect algebraic simplification when substituting a negative sum. For example, if the sum of roots is -5, they might write x² + 5x instead of x² - (-5)x which correctly simplifies to x² + 5x. The mistake usually happens when they incorrectly apply x² + (Sum of Roots)x instead of the correct x² - (Sum of Roots)x without accounting for the overall sign change.
✅ Correct Approach:
Always recall the standard form for a quadratic equation with roots α and β: x² - (α + β)x + αβ = 0.
  • First, calculate the sum of the roots (α + β) accurately.
  • Next, calculate the product of the roots (αβ) accurately.
  • Substitute these values into the standard form. Be especially careful with the negative sign preceding the sum of roots. If (α + β) is negative, ensure you write - (negative value), which will result in a positive term.
📝 Examples:
❌ Wrong:
Problem: Form a quadratic equation with roots -4 and 2.
Incorrect Steps:
Sum of roots (α + β) = -4 + 2 = -2
Product of roots (αβ) = (-4)(2) = -8
Incorrect equation formed: x² + (-2)x + (-8) = 0
Simplifies to: x² - 2x - 8 = 0
Reason for Error: The student incorrectly used x² + (Sum)x instead of x² - (Sum)x. The correct coefficient of 'x' should be - (-2) = +2, not -2.
✅ Correct:
Problem: Form a quadratic equation with roots -4 and 2.
Correct Steps:
Sum of roots (α + β) = -4 + 2 = -2
Product of roots (αβ) = (-4)(2) = -8
Using the standard form x² - (α + β)x + αβ = 0:
x² - (-2)x + (-8) = 0
Simplifies to: x² + 2x - 8 = 0
💡 Prevention Tips:
  • Memorize the Formula: Ensure you know the formula x² - (Sum)x + (Product) = 0 by heart, paying special attention to the minus sign.
  • Use Parentheses: Always enclose the sum and product of roots in parentheses when substituting them into the formula, especially if they are negative. For example, x² - (-2)x + (-8) = 0.
  • Separate Calculations: Calculate the sum and product of roots as distinct steps before substitution.
  • Double-Check Signs: After substitution, carefully review the signs of all terms in the equation.
  • For CBSE students: This is a straightforward concept, but minor sign errors are very common and can lead to loss of easy marks. Practice with negative roots is crucial.
CBSE_12th
Minor Unit Conversion

❌ Incorrect Sign Application for the Sum of Roots

Students frequently make a minor error by incorrectly applying the sign of the sum of roots when constructing the quadratic equation. While unit conversion is not directly applicable in forming quadratic equations from roots, this algebraic sign mistake can be seen as a 'misconversion' or misinterpretation of how the sum (S) contributes to the x² - Sx + P = 0 formula, particularly regarding the negative sign before 'S'. This leads to an incorrect coefficient for the 'x' term.
💭 Why This Happens:
This mistake primarily occurs due to:
  • Rote memorization: Students might memorize the formula x² - Sx + P = 0 without understanding its derivation from (x - α)(x - β) = 0, where α and β are the roots.
  • Careless algebraic slips: Even if the formula is known, a hurried calculation or substitution can lead to reversing the sign of the sum of roots.
  • Confusion with other formulas: Occasionally, students might confuse the sign convention with other algebraic expansions.
✅ Correct Approach:
The standard form for a quadratic equation with roots α and β is x² - (α + β)x + (αβ) = 0. Here, the sum of roots (S = α + β) always appears with a negative sign in front of it in the equation. The product of roots (P = αβ) appears with a positive sign.

Always remember: The coefficient of the 'x' term is the negative of the sum of the roots.
📝 Examples:
❌ Wrong:
Given roots: α = -2, β = 5
1. Calculate Sum of Roots (S): S = -2 + 5 = 3
2. Calculate Product of Roots (P): P = (-2) * 5 = -10

Incorrect Formation: A student might mistakenly write the equation as x² + Sx + P = 0 instead of x² - Sx + P = 0.
Substituting the values: x² + (3)x + (-10) = 0
Resulting in: x² + 3x - 10 = 0

This is incorrect because the coefficient of 'x' should be -3, not +3.
✅ Correct:
Given roots: α = -2, β = 5
  1. Calculate Sum of Roots (S): S = α + β = -2 + 5 = 3
  2. Calculate Product of Roots (P): P = αβ = (-2) * 5 = -10
  3. Apply the correct formula: x² - Sx + P = 0
  4. Substitute the values: x² - (3)x + (-10) = 0
  5. The correct quadratic equation is: x² - 3x - 10 = 0
💡 Prevention Tips:
  • Understand the Derivation (JEE & CBSE): Grasping that (x - α)(x - β) = x² - (α + β)x + αβ = 0 naturally clarifies the negative sign for the sum of roots.
  • Double-Check Signs (CBSE & JEE): After calculating the sum and product, always make a conscious effort to verify the sign of the middle term before finalizing the equation.
  • Consistent Practice (CBSE & JEE): Regularly solving problems with roots of varying signs (positive, negative, fractions) reinforces the correct application of the formula.
  • Verbalize the Formula: While writing, mentally or verbally state 'x squared, minus the sum of roots times x, plus the product of roots equals zero'.
CBSE_12th
Minor Formula

❌ Incorrect Sign for the Sum of Roots Term

A common mistake is using a positive sign for the 'sum of roots' term instead of the required negative sign when constructing a quadratic equation from its given roots. This leads to an incorrect quadratic equation.
💭 Why This Happens:
This error often arises from either superficial memorization of the formula or confusion with Vieta's formulas. Students might recall that if ax² + bx + c = 0, then the sum of roots is -b/a. However, when forming the equation in the monic form x² + (b/a)x + (c/a) = 0, the coefficient of x is actually -(sum of roots). Without understanding its derivation from (x - α)(x - β) = 0, students can easily swap the sign.
✅ Correct Approach:
The correct and universally accepted formula for forming a quadratic equation with roots α and β is:
x² - (α + β)x + αβ = 0
It is crucial to remember the negative sign before the sum of the roots (α + β). The term αβ (product of roots) always retains its natural sign.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3:
Incorrect attempt: x² + (2 + 3)x + (2 × 3) = 0
This results in: x² + 5x + 6 = 0 (which has roots -2 and -3)
✅ Correct:
If the roots are 2 and 3:
Sum of roots (α + β) = 2 + 3 = 5
Product of roots (αβ) = 2 × 3 = 6
Correct application: x² - (5)x + (6) = 0
The correct equation is: x² - 5x + 6 = 0
💡 Prevention Tips:
  • Memorize with Caution: Always commit the formula x² - (Sum of Roots)x + (Product of Roots) = 0 to memory, paying special attention to the minus sign.
  • Derivation Check: If unsure, quickly derive the formula from (x - α)(x - β) = 0, which expands to x² - (α + β)x + αβ = 0.
  • Practice with Negative Roots: Practice examples where roots are negative (e.g., -2 and -3) to solidify understanding of how the signs combine.
  • CBSE vs. JEE: This is a fundamental concept for both. A sign error will lead to completely wrong answers in both board exams and competitive tests.
CBSE_12th
Minor Conceptual

❌ Incorrect Sign Application in the Standard Quadratic Equation Form

A common conceptual error is misapplying the signs when using the standard formula to form a quadratic equation from its given roots. Students often forget that the term involving the sum of roots is subtracted, leading to an incorrect sign in the 'x' term of the final equation.
💭 Why This Happens:
This mistake primarily stems from a lack of thorough memorization of the correct standard form of a quadratic equation: x2 - (Sum of Roots)x + (Product of Roots) = 0. Instead, students might mistakenly use x2 + (Sum of Roots)x + (Product of Roots) = 0, or get confused when dealing with negative roots, leading to compounding sign errors during calculation of the sum and product.
✅ Correct Approach:
To form a quadratic equation with given roots α and β:
  • Step 1: Calculate the Sum of Roots (S) = α + β.
  • Step 2: Calculate the Product of Roots (P) = α * β.
  • Step 3: Substitute S and P into the standard equation form: x2 - Sx + P = 0.

Remember that the coefficient of the 'x' term is negative of the sum of the roots.

📝 Examples:
❌ Wrong:

Given Roots: 2 and 3

Student's Calculation:

  • Sum (S) = 2 + 3 = 5
  • Product (P) = 2 * 3 = 6

Incorrect Equation Formed: x2 + 5x + 6 = 0 (Mistake: Used '+' instead of '-' for the 'Sx' term).

✅ Correct:

Given Roots: 2 and 3

Correct Calculation:

  • Sum (S) = 2 + 3 = 5
  • Product (P) = 2 * 3 = 6

Correct Equation Formed: x2 - Sx + P = 0

                            x2 - (5)x + 6 = 0

                            x2 - 5x + 6 = 0

💡 Prevention Tips:
  • Memorize the Formula: Consistently use x2 - (Sum)x + (Product) = 0.
  • Double-Check Signs: Be extra careful when roots are negative; calculate S and P accurately first, then apply them to the formula.
  • Alternative Method (CBSE/JEE): If unsure, use the factor method: (x - α)(x - β) = 0. Expand this to verify your result from the sum/product method. This provides a cross-check.
CBSE_12th
Minor Calculation

❌ Sign Error in Sum of Roots Calculation

Students often make a sign error when calculating the sum of roots (α + β), especially when one or both roots are negative. While they correctly recall the general formula, x2 - (sum of roots)x + (product of roots) = 0, this calculation mistake leads to an incorrect sign for the 'x' term in the final quadratic equation.
💭 Why This Happens:
  • Carelessness: Rushing through basic arithmetic operations involving signed numbers.
  • Confusion: Misinterpreting the negative sign in - (α + β)x, sometimes incorrectly applying it to α + β instead of treating (α + β) as a single value.
  • Lack of Parentheses: Not using parentheses when adding or subtracting negative numbers, leading to errors.
✅ Correct Approach:
To avoid this, follow these steps meticulously:
  1. Identify Roots: Clearly note down the given roots, say α and β.
  2. Calculate Sum Carefully: Compute the sum of roots, S = α + β. Pay extra attention to signs.
  3. Calculate Product Carefully: Compute the product of roots, P = α * β.
  4. Substitute into Formula: Substitute the calculated S and P into the standard form: x2 - Sx + P = 0. The negative sign before Sx is critical.
📝 Examples:
❌ Wrong:
Incorrect Calculation:
Given roots are 2 and -5.
Sum of roots (S) = 2 + (-5) = 3 (Incorrect; should be -3)
Product of roots (P) = 2 * (-5) = -10
Using formula: x2 - (3)x + (-10) = 0
Result: x2 - 3x - 10 = 0
✅ Correct:
Correct Calculation:
Given roots are 2 and -5.
Sum of roots (S) = 2 + (-5) = -3
Product of roots (P) = 2 * (-5) = -10
Using formula: x2 - (-3)x + (-10) = 0
Result: x2 + 3x - 10 = 0
💡 Prevention Tips:
  • Write Explicitly: Always write down the calculated sum (S) and product (P) values explicitly before substituting them into the equation.
  • Use Parentheses: When dealing with negative numbers in sums (e.g., α + (-β)), use parentheses to group them clearly.
  • Double-Check: After calculating the sum, take a moment to re-verify the sign and value. This quick check can save marks.
  • Understand the Formula: Remember the formula is x2 - (Sum)x + (Product) = 0. The subtraction of the sum of roots is a fixed part of the formula.
CBSE_12th
Minor Approximation

❌ Premature Approximation of Irrational Roots or Coefficients

Students sometimes approximate irrational roots (e.g., √2, √3, π) to decimal values prematurely when calculating the sum (S) and product (P) of roots. This leads to non-exact, approximate coefficients in the final quadratic equation, which is generally not acceptable in JEE Advanced unless the question explicitly asks for an approximate answer or a rounded form.
💭 Why This Happens:
This mistake commonly occurs due to:
  • A desire to simplify calculations involving irrational numbers.
  • Misunderstanding the precision required in JEE Advanced, where exact forms are usually expected.
  • Habit from other contexts (like physics or chemistry numericals) where decimal approximations are frequently used.
✅ Correct Approach:
Always perform calculations for the sum (S = α + β) and product (P = αβ) of roots using their exact irrational forms. The coefficients of the quadratic equation x² - Sx + P = 0 must also be kept in their exact forms (integers, rational numbers, or expressions involving irrational numbers) unless specified otherwise. This ensures the derived equation is perfectly accurate.
📝 Examples:
❌ Wrong:
If the roots are α = 1 + √3 and β = 2, a student might approximate √3 ≈ 1.732.
Then:
Sum (S) ≈ (1 + 1.732) + 2 = 2.732 + 2 = 4.732
Product (P) ≈ (1 + 1.732) * 2 = 2.732 * 2 = 5.464
The resulting equation: x² - 4.732x + 5.464 = 0. This equation contains approximate coefficients and will likely be marked incorrect in JEE Advanced.
✅ Correct:
Given the same roots: α = 1 + √3 and β = 2.
Correct calculation:
Sum (S) = (1 + √3) + 2 = 3 + √3
Product (P) = (1 + √3) * 2 = 2 + 2√3
The quadratic equation is x² - Sx + P = 0.
Substituting S and P: x² - (3 + √3)x + (2 + 2√3) = 0. This is the exact and expected form in competitive exams like JEE Advanced.
💡 Prevention Tips:
  • JEE Advanced Specific: Always assume that exact values for coefficients are required unless the question explicitly asks for a decimal approximation or rounding to a specific number of decimal places.
  • Work with Exact Forms: Keep irrational numbers in their symbolic form (e.g., √2, √3, π) throughout calculations for sum and product.
  • Final Check: Before marking your answer, ensure that all coefficients of the quadratic equation are in their simplest exact form.
JEE_Advanced
Minor Conceptual

❌ Assuming Real/Rational Coefficients by Default

Students often incorrectly assume quadratic equations must have real or rational coefficients, even when given roots are not conjugate pairs. This leads to errors when forming equations or determining feasibility under specific coefficient constraints.
💭 Why This Happens:
Frequent exposure to real-coefficient problems instills this default assumption, making students overlook that equations can have complex/irrational coefficients if not specified.
✅ Correct Approach:

  1. Use the standard formula $x^2 - (alpha + eta)x + alphaeta = 0$ with the given roots $alpha$ and $eta$.

  2. JEE Advanced Tip: Always check for explicit conditions on coefficients (e.g., "form a quadratic equation with real coefficients").

    • If "real coefficients" is stated, and one root is complex ($a+ib, b
      eq 0$), the other root must be its conjugate ($a-ib$). If the given roots do not form this pair, then no such quadratic equation with real coefficients can be formed.

    • If no coefficient condition is stated, the equation formed directly using the given roots (even if non-conjugate) is perfectly valid, and its coefficients may be complex or irrational.



📝 Examples:
❌ Wrong:
Suppose the roots are given as $alpha = 2$ and $eta = 3+i$.

Wrong:

Assuming real coefficients are required, a student might incorrectly conclude that no such equation can be formed because $3-i$ is not a given root. This is a conceptual mistake if "real coefficients" was not a stated condition in the problem.
✅ Correct:
Consider the roots $alpha = 2$ and $eta = 3+i$.

Correct:



  1. Sum of roots: $alpha + eta = 2 + (3+i) = 5+i$

  2. Product of roots: $alphaeta = 2(3+i) = 6+2i$

  3. The quadratic equation is: $x^2 - (5+i)x + (6+2i) = 0$.


This is a perfectly valid quadratic equation with complex coefficients. No condition on coefficients was specified, so direct application of the formula is correct.

💡 Prevention Tips:

  • Read Carefully: Always note if there are explicit conditions on the nature of coefficients (e.g., "real coefficients", "rational coefficients").

  • Conjugate Pairs Rule: Remember that complex (or irrational) roots occur in conjugate pairs only if the quadratic equation has real (or rational) coefficients, respectively. This is a consequence, not a default assumption.

  • Direct Formula Application: If no conditions on coefficients are given, directly apply $x^2 - ( ext{Sum of Roots})x + ( ext{Product of Roots}) = 0$ using the exact given roots.

JEE_Advanced
Minor Sign Error

❌ Sign Error in the Sum of Roots Term

Students frequently make a sign error when incorporating the sum of roots into the quadratic equation. Instead of using x2 - (sum of roots)x + (product of roots) = 0, they incorrectly use x2 + (sum of roots)x + (product of roots) = 0, especially when the sum of roots turns out to be negative.
💭 Why This Happens:
This error primarily stems from a conceptual confusion or a lapse in memory regarding the standard form of a quadratic equation derived from its roots. Many students recall that the coefficient of 'x' is related to the sum of roots, but forget the crucial negative sign that precedes it. Carelessness during calculations involving negative roots also contributes significantly to this mistake. For JEE Advanced, where precision is paramount, such minor errors can be very costly.
✅ Correct Approach:
The correct approach involves strictly adhering to the fundamental formula for forming a quadratic equation with given roots α and β:
x2 - (α + β)x + (αβ) = 0
Here, (α + β) is the sum of the roots, and (αβ) is the product of the roots. Always substitute the calculated sum and product carefully, paying close attention to their signs.
📝 Examples:
❌ Wrong:
Given roots: α = -3, β = 2
Sum of roots (α + β): -3 + 2 = -1
Product of roots (αβ): (-3)(2) = -6
Incorrect Equation: x2 + (-1)x + (-6) = 0 → x2 - x - 6 = 0 (Wait! This example is correct for sum of roots being -1. I need to make the wrong one actually wrong in the x term sign.)
Let's re-evaluate for a clear 'wrong' example.
Incorrect Equation (Common Mistake): If a student *forgets* the minus sign in the formula and uses x2 + (sum)x + (product) = 0 directly:
x2 + (-1)x + (-6) = 0 is still x2 - x - 6 = 0. This is the correct equation.
A true sign error would be:
x2 + (sum of roots)x + (product of roots) = 0
If sum = 1, product = -6. Wrong equation: x2 + x - 6 = 0.
Let's pick roots where sum is positive, and the mistake would be putting a plus sign when it should be minus for the sum term.

Given roots: α = 5, β = -3
Sum of roots (α + β): 5 + (-3) = 2
Product of roots (αβ): (5)(-3) = -15
Wrong Equation (Common Mistake): Using x2 + (sum)x + (product) = 0 directly without the leading negative sign for sum:
x2 + (2)x + (-15) = 0
x2 + 2x - 15 = 0
✅ Correct:
Given roots: α = 5, β = -3
Sum of roots (α + β): 5 + (-3) = 2
Product of roots (αβ): (5)(-3) = -15
Correct Equation: Using x2 - (sum of roots)x + (product of roots) = 0
x2 - (2)x + (-15) = 0
x2 - 2x - 15 = 0
💡 Prevention Tips:
  • Memorize the Formula Precisely: Commit x2 - (Sum of Roots)x + (Product of Roots) = 0 to memory, paying special attention to the minus sign before the sum term.
  • Vieta's Relations Check: Always cross-verify with Vieta's relations: if ax2 + bx + c = 0, then α + β = -b/a and αβ = c/a. This ensures the sign of the 'x' coefficient is correct.
  • Substitute with Parentheses: When substituting the sum and product into the formula, especially if they are negative, use parentheses to avoid sign errors, e.g., x2 - (-5)x + (-6) = 0.
  • Practice with Negative Roots: Work through multiple examples where roots are negative or have mixed signs to solidify understanding.
JEE_Advanced
Minor Unit Conversion

❌ Incorrect Sign Application for Sum of Roots

Students frequently misapply the negative sign associated with the sum of roots `(α+β)` when forming a quadratic equation. Instead of using `x^2 - (α+β)x + αβ = 0`, they might incorrectly use `x^2 + (α+β)x + αβ = 0`, leading to an erroneous coefficient for the 'x' term. This is a common minor error in 'converting' the numerical properties of roots into the correct equation structure.
💭 Why This Happens:
This error often stems from rote memorization without understanding the derivation of the formula from `(x-α)(x-β) = 0`. The fundamental algebraic expansion includes a `-(α+β)x` term. Students might confuse the direct sum with its placement in the equation. In the context of 'Unit Conversion understanding', this can be seen as a misinterpretation of how numerical values (the sum of roots) are 'converted' or 'translated' into their corresponding coefficient in the standard quadratic form, leading to a sign 'conversion' error.
✅ Correct Approach:
Always recall the standard form for a quadratic equation with given roots `α` and `β`: x2 - (α+β)x + αβ = 0. Calculate the sum of roots `(α+β)` and the product of roots `(αβ)` first, paying careful attention to their signs. Then, substitute these values into the formula, ensuring the negative sign before the sum of roots term is correctly applied.
📝 Examples:
❌ Wrong:
Given roots: α = 2, β = -5.
Sum of roots: α+β = 2 + (-5) = -3.
Product of roots: αβ = 2 * (-5) = -10.
Incorrect approach: Student forms the equation as x2 + (α+β)x + αβ = 0.
Substituting values: x2 + (-3)x + (-10) = 0
Resulting in: x2 - 3x - 10 = 0
✅ Correct:
Given roots: α = 2, β = -5.
Sum of roots: α+β = 2 + (-5) = -3.
Product of roots: αβ = 2 * (-5) = -10.
Correct approach: Using the formula x2 - (α+β)x + αβ = 0.
Substituting values carefully: x2 - (-3)x + (-10) = 0
Resulting in: x2 + 3x - 10 = 0
💡 Prevention Tips:
  • CBSE/JEE Tip: Always explicitly write down the formula: x2 - (Sum of Roots)x + (Product of Roots) = 0.
  • Calculate the sum (α+β) and product (αβ) of roots separately, double-checking signs.
  • When substituting, use parentheses around the calculated sum and product to avoid sign errors: x2 - (calculated sum)x + (calculated product) = 0.
  • Important: In pure mathematics, roots are typically dimensionless numerical values. While 'unit conversion' isn't directly applicable here, errors often arise from misinterpreting numerical signs or magnitudes during the 'conversion' of root properties into equation coefficients. Pay close attention to these numerical details.
JEE_Advanced
Minor Formula

❌ Ignoring the General Form Constant 'k'

Students often recall the formula for forming a quadratic equation with roots α and β as x² - (α+β)x + αβ = 0. While this gives a valid quadratic equation, it is not the most general form. The mistake is to forget or ignore the multiplicative constant 'k' that can precede the entire expression, especially in JEE Advanced problems where additional conditions might be given to determine 'k'.
💭 Why This Happens:
This mistake primarily stems from an oversimplification of the formula learned in earlier stages. Students tend to remember the simplest form that yields a quadratic equation, rather than the complete general form. Lack of understanding that k(ax² + bx + c) = 0 represents the same roots as ax² + bx + c = 0, but describes a different equation (with scaled coefficients), is a contributing factor. For JEE Advanced, this 'k' factor is frequently crucial for unique identification of the quadratic.
✅ Correct Approach:
The correct and most general form for a quadratic equation with roots α and β is k[x² - (α+β)x + αβ] = 0, where k is any non-zero real constant.
  • If no additional information is provided, k=1 is usually assumed, giving the simplest equation.
  • However, if the question specifies the leading coefficient, a point the quadratic passes through, or other conditions, 'k' must be determined using that information.
📝 Examples:
❌ Wrong:
Problem: Form a quadratic equation with roots 2 and 3, such that the leading coefficient is 4.
Wrong Approach: Sum of roots (S) = 2+3 = 5. Product of roots (P) = 2*3 = 6.
Equation is x² - 5x + 6 = 0. (Here, the leading coefficient is 1, not 4).
✅ Correct:
Problem: Form a quadratic equation with roots 2 and 3, such that the leading coefficient is 4.
Correct Approach:
Sum of roots (S) = 2+3 = 5.
Product of roots (P) = 2*3 = 6.
General form: k[x² - (S)x + P] = 0
k[x² - 5x + 6] = 0
To have a leading coefficient of 4, the coefficient of x² must be 4. In our general form, the coefficient of x² is k. So, we set k=4.
The correct equation is 4(x² - 5x + 6) = 0, which simplifies to 4x² - 20x + 24 = 0.
💡 Prevention Tips:
Always start with the general form: k[x² - (sum of roots)x + (product of roots)] = 0.
Carefully read the question for any additional constraints (e.g., leading coefficient, passing through a point) that might help determine the value of 'k'.
Understand that k scales the entire equation without changing its roots. For JEE Advanced, neglecting 'k' can lead to incorrect answers when specific properties of the quadratic are given.
JEE_Advanced
Minor Calculation

❌ Sign Errors and Algebraic Simplification Mistakes in Sum/Product of Roots

Students frequently make minor calculation errors while determining the sum (α+β) and product (αβ) of the given roots. These often involve incorrect handling of negative signs, fractions, or radicals during addition, subtraction, or multiplication, leading to an incorrect quadratic equation of the form x² - (α+β)x + αβ = 0.
💭 Why This Happens:
This mistake primarily stems from a lack of carefulness during calculations.
  • Haste: Rushing through steps, especially in a time-bound exam like JEE Advanced.
  • Weak algebraic manipulation: Basic errors in adding/multiplying integers, fractions, or irrational numbers.
  • Sign confusion: Misinterpreting signs when combining terms or multiplying two negative numbers.
While seemingly minor, such errors can result in a completely wrong final answer.
✅ Correct Approach:
Always calculate the sum and product of roots step-by-step with utmost precision.
  • Sum of roots (α+β): Carefully add the given roots, paying close attention to their signs.
  • Product of roots (αβ): Carefully multiply the given roots, again, being mindful of signs and basic algebraic rules.
Once α+β and αβ are correctly found, substitute them into the general form: x² - (α+β)x + αβ = 0. For JEE, sometimes coefficients 'a' are implied, so the general form is a(x² - (α+β)x + αβ) = 0.
📝 Examples:
❌ Wrong:
Given roots α = -3 and β = 2/3.
Wrong Calculation:
Sum (α+β) = -3 + 2/3 = (-9 + 2)/3 = -7/3 (Correct here)
Product (αβ) = -3 * 2/3 = -6/3 = -2 (Correct here)
Equation: x² - (-7/3)x + (-2) = 0 => 3x² - 7x - 6 = 0 (Sign error in the coefficient of x - should be +7x)
✅ Correct:
Given roots α = -3 and β = 2/3.
Correct Calculation:
Sum (α+β) = -3 + 2/3 = (-9 + 2)/3 = -7/3
Product (αβ) = -3 * 2/3 = -2
Using the formula x² - (α+β)x + αβ = 0:
x² - (-7/3)x + (-2) = 0
x² + (7/3)x - 2 = 0
Multiplying by 3 to clear the denominator:
3x² + 7x - 6 = 0
💡 Prevention Tips:
  • Double-check: Always re-verify your sum and product calculations before forming the equation.
  • Write down all steps: Avoid mental calculations, especially with negative numbers, fractions, or complex expressions.
  • Be mindful of signs: This is the most common pitfall. 'Minus of minus' is 'plus', etc.
  • Practice: Regular practice with various types of roots (integers, fractions, radicals) strengthens basic algebraic proficiency, crucial for JEE Advanced.
JEE_Advanced
Important Other

❌ Neglecting the Arbitrary Constant 'k' in Equation Formation

Students frequently form the quadratic equation using the sum (S) and product (P) of roots as x2 - Sx + P = 0, often overlooking the general form which includes an arbitrary non-zero constant 'k'. This oversight can lead to an incomplete or incorrect solution, especially in JEE Main problems where additional conditions might be provided to determine 'k'.
💭 Why This Happens:
  • Many introductory problems implicitly assume k=1, leading students to believe it's always the case.
  • A lack of clear understanding that ax2 + bx + c = 0 and k(ax2 + bx + c) = 0 represent the same roots for any k ≠ 0.
  • Over-reliance on rote memorization of the basic formula without grasping its underlying general nature.
✅ Correct Approach:
The general form of a quadratic equation with roots α and β is k(x2 - (α + β)x + αβ) = 0, where k ≠ 0 is an arbitrary non-zero constant. This 'k' is critical when the problem specifies a leading coefficient or provides an additional point through which the quadratic passes, allowing for the determination of a unique equation.
📝 Examples:
❌ Wrong:

Problem: Form a quadratic equation with roots 2 and 3.

Incorrect Approach:

Sum of roots (S) = 2 + 3 = 5

Product of roots (P) = 2 * 3 = 6

Equation: x2 - 5x + 6 = 0



This equation is correct if k=1, but it's not the general form and doesn't account for other valid equations like 2x2 - 10x + 12 = 0.
✅ Correct:

Problem: Form a quadratic equation with roots 2 and 3 that passes through the point (1, -2).

Correct Approach:

Given roots α = 2, β = 3.

Sum of roots (S) = α + β = 2 + 3 = 5

Product of roots (P) = αβ = 2 * 3 = 6



The general form of the quadratic equation is: k(x2 - Sx + P) = 0

Substitute S and P: k(x2 - 5x + 6) = 0



Now, use the additional condition that the equation passes through (1, -2):

Substitute x=1 and y=-2 into the general equation:

k((1)2 - 5(1) + 6) = -2

k(1 - 5 + 6) = -2

k(2) = -2

k = -1



Therefore, the specific quadratic equation is: -1(x2 - 5x + 6) = 0, which simplifies to -x2 + 5x - 6 = 0.



Without including 'k', it would be impossible to determine this unique equation from the given conditions.
💡 Prevention Tips:
  • Always use the general form: Start with k(x2 - Sx + P) = 0.
  • Identify when 'k' is needed:
    • For CBSE Board Exams, x2 - Sx + P = 0 is often sufficient unless a specific leading coefficient or a point is mentioned.
    • For JEE Main/Advanced, 'k' is crucial. Look for conditions like: 'the quadratic passes through a point (a, b)', 'the leading coefficient is c', or 'find *the* specific quadratic equation'.
  • Double-check problem wording: Distinguish between 'a quadratic equation' (where 'k' can be any non-zero value, often implicitly 1) and 'the quadratic equation' (which implies a unique 'k' must be found).
JEE_Main
Important Approximation

❌ Approximating Irrational or Complex Roots Prematurely

Students often make the critical error of approximating irrational roots (e.g., √3 ≈ 1.732) or complex roots (if they involve irrational parts) to decimal values before computing their sum and product to form the quadratic equation. This leads to an approximate equation with non-integer/non-exact coefficients, which is incorrect for JEE Main where exact solutions are expected.
💭 Why This Happens:
This mistake stems from a misunderstanding that all calculations must be exact until the final simplified form. Students might be used to using calculators for decimal approximations in other contexts or mistakenly believe that simplifying means converting to decimals. They fail to recognize that substituting approximate values will yield an approximate quadratic equation, not the precise one required.
✅ Correct Approach:
Always use the exact algebraic forms of the roots (irrational or complex) when calculating the sum of roots (α+β) and the product of roots (αβ). Substitute these exact values into the standard formula: x² - (α+β)x + αβ = 0. The coefficients of the resulting quadratic equation must be exact, typically integers or simple fractions, unless the problem explicitly states otherwise (which is rare for equation formation).
📝 Examples:
❌ Wrong:
Given roots are α = 2 + √5 and β = 2 - √5.
Student approximates √5 ≈ 2.236.
α ≈ 2 + 2.236 = 4.236
β ≈ 2 - 2.236 = -0.236
Sum (α+β) ≈ 4.236 - 0.236 = 4.000
Product (αβ) ≈ (4.236)(-0.236) ≈ -1.000656 (approximated to -1)
Approximate Equation: x² - 4x - 1 = 0. While this looks correct, the intermediate approximation of √5 makes it conceptually flawed.
✅ Correct:
Given roots are α = 2 + √5 and β = 2 - √5.
Step 1: Calculate the exact sum of roots.
α+β = (2 + √5) + (2 - √5) = 2 + 2 + √5 - √5 = 4
Step 2: Calculate the exact product of roots.
αβ = (2 + √5)(2 - √5) = 2² - (√5)² = 4 - 5 = -1
Step 3: Form the quadratic equation using exact values.
x² - (α+β)x + αβ = 0
x² - (4)x + (-1) = 0
x² - 4x - 1 = 0. This is the exact and correct quadratic equation.
💡 Prevention Tips:
  • Never approximate irrational or complex numbers to decimals when forming quadratic equations, unless specifically instructed in a rare scenario.
  • Always perform algebraic operations on the exact forms of the roots (e.g., (1+√3)² or (2+3i)(2-3i)).
  • Understand that JEE Main questions typically demand exact answers for coefficients of quadratic equations.
  • Double-check calculations for sum and product of roots to avoid arithmetic errors, especially with signs and squaring terms like (a+b√c).
JEE_Main
Important Conceptual

❌ Sign Errors in the Sum of Roots Term

Students frequently make sign errors when forming a quadratic equation using the sum and product of its roots. The most common mistake is incorrectly applying the negative sign in the general formula x² - (sum of roots)x + (product of roots) = 0, especially when the sum of roots itself is a negative value. They might mistakenly write x² + (sum of roots)x + (product of roots) = 0, leading to an incorrect coefficient for the 'x' term.
💭 Why This Happens:
This conceptual error often stems from a lack of thorough understanding of the derivation of the quadratic equation from its roots (i.e., from (x - α)(x - β) = 0). When expanded, this yields x² - (α + β)x + αβ = 0. The negative sign before the sum of roots, -(α + β), is crucial. Students might overlook this standard form and simply substitute the calculated sum and product without considering the inherent sign in the formula. Rushing calculations or a superficial understanding of Vieta's formulas also contributes.
✅ Correct Approach:
Always remember and strictly adhere to the standard form of a quadratic equation when roots (α and β) are given:
x² - (α + β)x + αβ = 0.
First, accurately calculate the sum of the roots (S = α + β) and the product of the roots (P = αβ), paying close attention to their individual signs. Then, substitute these values into the formula, being particularly careful with the negative sign that precedes the sum of roots term. Think of the coefficient of 'x' as '-(sum of roots)'.
📝 Examples:
❌ Wrong:
Given Roots: α = -2, β = 3
Step 1: Calculate Sum (S) = -2 + 3 = 1
Step 2: Calculate Product (P) = (-2) * 3 = -6
Student's Incorrect Equation: Mistakenly using x² + Sx + P = 0 instead of the correct form.
x² + (1)x + (-6) = 0
x² + x - 6 = 0 (Incorrect quadratic equation)
✅ Correct:
Given Roots: α = -2, β = 3
Step 1: Calculate Sum (S) = -2 + 3 = 1
Step 2: Calculate Product (P) = (-2) * 3 = -6
Correct Equation using x² - (S)x + P = 0:
x² - (1)x + (-6) = 0
x² - x - 6 = 0 (Correct quadratic equation)
💡 Prevention Tips:
  • Memorize the Formula: Ensure you know the formula x² - (Sum)x + (Product) = 0 perfectly. The minus sign before the sum is critical.
  • Separate Calculations: Always calculate the sum of roots (S) and product of roots (P) as distinct steps.
  • Parentheses for Substitution: When substituting S and P into the formula, especially if they are negative, use parentheses: x² - (S)x + (P) = 0. This helps prevent sign errors.
  • Self-Check: After forming the equation, quickly find its roots using the quadratic formula or factorization to verify if they match the given roots.
  • Practice: Solve a variety of problems with both positive and negative roots to build confidence and reinforce the correct application of signs.
JEE_Advanced
Important Calculation

❌ Sign Errors in Calculation of Sum/Product of Roots and Formula Substitution

A frequent calculation mistake involves incorrect handling of signs when determining the sum (α+β) and product (αβ) of given roots, or during their substitution into the standard quadratic equation formula. This is particularly prevalent with negative or fractional roots.
💭 Why This Happens:
This error often stems from a combination of haste, neglecting basic arithmetic rules for signed numbers, and overlooking the negative sign in the general formula: x² - (sum of roots)x + (product of roots) = 0. Under exam pressure, students may mistakenly use + (sum of roots)x.
✅ Correct Approach:
Always meticulously calculate the sum and product of the roots, paying close attention to their signs. Then, substitute these values carefully into the formula x² - (α + β)x + αβ = 0. Remember the crucial negative sign before the sum of roots.
📝 Examples:
❌ Wrong:
Given roots are 2 and -5.
Sum (α+β) = 2 + (-5) = -3
Product (αβ) = 2 * (-5) = -10

Wrong Calculation/Substitution:
If a student mistakenly calculates sum as 3 or uses `x² + (sum)x + product = 0`, they might write:
x² + 3x - 10 = 0 (Incorrect sum used)
OR
x² - 3x + 10 = 0 (Incorrect product sign used)
✅ Correct:
Given roots are 2 and -5.
Sum (α+β) = 2 + (-5) = -3
Product (αβ) = 2 * (-5) = -10

Correct Approach:
Using the formula: x² - (α + β)x + αβ = 0
Substitute carefully: x² - (-3)x + (-10) = 0
Simplify: x² + 3x - 10 = 0
Note the change from -(-3) to +3.
💡 Prevention Tips:
  • Explicitly Write Down: Always write down the calculated (α+β) and αβ with their correct signs before substitution.
  • Use Parentheses: When substituting negative values, especially into -(α+β), use parentheses: -(-3).
  • Double-Check Formula: Confirm that you are using x² - (sum)x + (product) = 0, not x² + (sum)x + (product) = 0.
  • JEE Advanced Tip: For crucial problems, if time permits, substitute one of the given roots back into your final quadratic equation to verify if it satisfies the equation. This can catch calculation errors.
  • Practice with Variety: Solve problems involving various types of roots (negative, fractional, irrational) to build accuracy in arithmetic and sign handling.
JEE_Advanced
Important Formula

❌ Miscalculating Sum/Product of Roots due to Neglecting Conjugate Pairs (JEE Advanced Focus)

Students often make the critical error of neglecting the 'conjugate root theorem' when forming quadratic equations with real coefficients. If one root given is irrational (e.g., a + √b) or complex (e.g., a + ib), they forget that the other root must be its conjugate (a - √b or a - ib, respectively). This leads to incorrect calculations of the sum and product of roots, consequently forming an erroneous quadratic equation. This is particularly relevant for JEE Advanced where such implicit conditions are frequently tested.
💭 Why This Happens:
  • Lack of Conceptual Clarity: Insufficient understanding of the property of roots for quadratic equations specifically with real coefficients.
  • Overlooking Implicit Conditions: Failing to recognize that if a problem provides only one irrational or complex root for a quadratic equation, the existence of real coefficients implies the presence of its conjugate.
  • Direct Substitution Error: Directly assuming the given single root as 'the sum' or 'the product' without deducing the second root.
✅ Correct Approach:
  • Identify Coefficient Nature: Always check if the quadratic equation is stated or implied to have real coefficients.
  • Apply Conjugate Root Theorem: If one root (α) is irrational (a + √b) or complex (a + ib), and the coefficients are real, then the other root (β) must be its conjugate (a - √b or a - ib, respectively).
  • Calculate Correct Sum & Product: Once both roots (α and β) are correctly identified, calculate their sum (α + β) and product (αβ).
  • Form the Equation: Use the standard formula: x² - (α + β)x + αβ = 0.
📝 Examples:
❌ Wrong:

Problem: Form a quadratic equation with real coefficients if one root is 3 + √5.

  • Student's Incorrect Approach:
  • Assumes the given root (α = 3 + √5) is the only root, or mistakenly uses it for both roots without considering the conjugate.
  • Calculates: Sum (S) = 3 + √5, Product (P) = (3 + √5)² = 9 + 5 + 6√5 = 14 + 6√5.
  • Forms: x² - (3 + √5)x + (14 + 6√5) = 0. (Note: This equation will have irrational coefficients, violating the 'real coefficients' condition if only one root is given and it's irrational).
✅ Correct:

Problem: Form a quadratic equation with real coefficients if one root is 3 + √5.

  • Correct Approach:
  • Since coefficients are real and one root is α = 3 + √5 (irrational), the other root must be its conjugate: β = 3 - √5.
  • Sum of roots (S): α + β = (3 + √5) + (3 - √5) = 6.
  • Product of roots (P): αβ = (3 + √5)(3 - √5) = 3² - (√5)² = 9 - 5 = 4.
  • Using the formula x² - Sx + P = 0:
  • The correct quadratic equation is x² - 6x + 4 = 0.
💡 Prevention Tips:
  • JEE Advanced Insight: Always be alert to the phrasing 'real coefficients' when dealing with irrational or complex roots. This is a common trap designed to test deeper understanding.
  • Mind Map for Conjugates: If one root is given as a + √b or a + ib, immediately write down the other root as its conjugate a - √b or a - ib, respectively, before proceeding to calculate sum and product.
  • Self-Check: After forming the equation, quickly verify if its coefficients are indeed real, especially if the problem specified 'real coefficients'.
JEE_Advanced
Important Unit Conversion

❌ Misapplying 'Unit Conversion' Concepts to Numerical Roots

A common conceptual error, particularly for students who might overthink or confuse algebraic contexts with physics/chemistry problems, is to incorrectly assume that the numerical values given as roots of a quadratic equation possess physical units or require 'conversion' before being used in the standard formulas. Roots, in the context of forming a quadratic equation, are purely abstract, dimensionless numbers.
💭 Why This Happens:
This mistake stems from a fundamental misunderstanding of what a 'root' represents in algebra—it's a numerical value for the variable, not a physical quantity. Students might:
  • Lack clarity on the dimensionless nature of roots in abstract mathematics.
  • Confuse algebraic problems with applied problems in physics or chemistry where unit consistency is crucial.
  • Over-analyze simple algebraic statements, attempting to introduce a layer of complexity (like unit conversion) where none exists.
✅ Correct Approach:
The roots of a quadratic equation (α and β) are simply numerical values. To form the quadratic equation, you directly use these numbers to calculate their sum and product. The standard formula is: x2 - (Sum of roots)x + (Product of roots) = 0. No unit conversion is ever required or relevant for the numerical values of roots in this algebraic context.
📝 Examples:
❌ Wrong:
If roots are given as 2 and 3. A student might incorrectly ponder: 'What if one root represents 2 meters and the other 3 seconds? How do I make them 'unit consistent' before adding them?' This line of thought is fundamentally incorrect for forming a quadratic equation from given roots.
✅ Correct:
Given roots α = 2 and β = 3 (these are just numbers).
  • Sum of roots: α + β = 2 + 3 = 5
  • Product of roots: αβ = 2 × 3 = 6
The quadratic equation is simply x2 - 5x + 6 = 0. No units were considered or converted.
💡 Prevention Tips:
  • Understand the Basics: Always remember that in algebraic contexts, especially for 'formation of quadratic equations with given roots', roots are treated as pure numerical values (real, complex, rational, irrational). They do not carry units.
  • Contextual Awareness: Distinguish between abstract mathematical problems and applied problems (e.g., word problems in physics/chemistry) where quantities might have units. For JEE Advanced, if units are relevant, they will be explicitly stated within a word problem, and the 'formation' part would still use the numerical values derived.
  • Stick to the Formula: The formulas for sum (α + β) and product (αβ) operate purely on the numerical values. There's no scope for 'unit conversion' in this process.
  • JEE Advanced Note: While JEE Advanced tests conceptual depth, it will not trick you by subtly implying units on abstract roots. Focus on the mathematical properties of the numbers themselves.
JEE_Advanced
Important Other

❌ Incorrect Sign for the Sum of Roots Term

Students frequently make sign errors when substituting the sum of roots into the general form of a quadratic equation. Specifically, they often forget or incorrectly apply the negative sign before the sum of roots term, leading to an incorrect coefficient for 'x'.
💭 Why This Happens:
  • Forgetting the Standard Formula: The correct standard form is x2 - (α + β)x + αβ = 0. Students often mistakenly use x2 + (α + β)x + αβ = 0.
  • Careless Substitution: If the sum of roots (α + β) itself is negative (e.g., -5), students might incorrectly write - (-5)x as -5x instead of +5x.
  • Lack of Conceptual Understanding: Not fully grasping that the coefficient of 'x' (b/a) is related to the negative of the sum of roots, -(α + β).
✅ Correct Approach:
To form a quadratic equation with roots α and β, follow these steps:
  1. Calculate the sum of the roots (α + β).
  2. Calculate the product of the roots (αβ).
  3. Substitute these values into the standard formula:
    x2 - (Sum of Roots)x + (Product of Roots) = 0
    Pay meticulous attention to the negative sign preceding the sum of roots.
    For CBSE & JEE, this formula is fundamental.
📝 Examples:
❌ Wrong:
Given roots: -3 and 2
Sum of roots (α + β): -3 + 2 = -1
Product of roots (αβ): (-3) * 2 = -6
Wrong Equation: x2 + (-1)x + (-6) = 0
x2 - x - 6 = 0
(Mistake: Used '+' before the sum of roots instead of '-')
✅ Correct:
Given roots: -3 and 2
Sum of roots (α + β): -3 + 2 = -1
Product of roots (αβ): (-3) * 2 = -6
Correct Equation: x2 - (-1)x + (-6) = 0
x2 + x - 6 = 0
(Note: This can be verified by factoring as (x+3)(x-2)=0, which expands to x2+x-6=0)
💡 Prevention Tips:
  • Memorize Accurately: Commit the formula x2 - (α + β)x + αβ = 0 to memory precisely, including the negative sign.
  • Separate Calculations: Always calculate (α + β) and (αβ) as distinct values before substituting them into the equation.
  • Double-Check Signs: When substituting, explicitly write out the sign multiplication. For instance, if (α + β) = -5, write - (-5)x which simplifies to +5x.
  • Verification: For JEE Advanced, quickly verify by expanding the factored form (x - α)(x - β) = 0 to cross-check your derived equation.
JEE_Advanced
Important Approximation

❌ Premature Approximation of Irrational or Complex Roots

Students often approximate irrational numbers (like √2, √3) or complex numbers (e.g., replacing i with a decimal approximation if they mistakenly think it's a real number, or rounding parts of complex numbers) before forming the quadratic equation. This leads to an incorrect sum and product of roots, resulting in an inaccurate equation, especially when exact values are required in JEE Advanced.
💭 Why This Happens:
This mistake stems from a lack of understanding of the precision required in competitive exams. Students might be uncomfortable with algebraic manipulation involving surds or imaginary units, or they might rely on calculators (which are generally not allowed) to simplify calculations, thereby introducing rounding errors. Sometimes, it's a conceptual oversight, failing to recognize that roots must be handled in their exact form.
✅ Correct Approach:
Always work with the exact values of roots, whether they are irrational (involving surds) or complex (involving i). Perform all algebraic operations for finding the sum (α + β) and product (αβ) of roots precisely. Only if the problem explicitly asks for an approximate answer should any rounding occur, and that too, only at the final step. For JEE Advanced, exact answers are almost always expected.
📝 Examples:
❌ Wrong:
Problem: Form a quadratic equation whose roots are 1 + √3 and 2 - √3.
Student's Wrong Approach:
Approximate √3 ≈ 1.732.
Roots become 1 + 1.732 = 2.732 and 2 - 1.732 = 0.268.
Sum (S) ≈ 2.732 + 0.268 = 3.000.
Product (P) ≈ 2.732 × 0.268 ≈ 0.732176.
Quadratic Equation: x² - 3x + 0.732 = 0. (Incorrect)
✅ Correct:
Problem: Form a quadratic equation whose roots are 1 + √3 and 2 - √3.
Correct Approach:
Let the roots be α = 1 + √3 and β = 2 - √3.
1. Sum of roots (S):
   S = α + β = (1 + √3) + (2 - √3) = 1 + 2 + √3 - √3 = 3.
2. Product of roots (P):
   P = αβ = (1 + √3)(2 - √3) = 1(2) - 1(√3) + √3(2) - √3(√3)
   P = 2 - √3 + 2√3 - 3 = √3 - 1.
3. Form the equation: x² - Sx + P = 0
   x² - 3x + (√3 - 1) = 0. (Correct exact equation)
💡 Prevention Tips:
  • Prioritize Exactness: Always work with surds, fractions, and the imaginary unit 'i' in their exact forms until the final required answer.
  • Algebraic Simplification: Practice simplifying expressions involving irrational and complex numbers algebraically. This builds confidence and reduces the temptation to approximate.
  • JEE Advanced Expectation: Understand that JEE Advanced problems typically demand precise, exact answers unless an approximation is explicitly requested (e.g., 'to two decimal places').
  • Check your Work: After forming the equation, quickly substitute one of the original roots to ensure it satisfies the equation, verifying your sum and product calculations.
JEE_Advanced
Important Sign Error

❌ Sign Error in the Coefficient of the 'x' Term

A common mistake in forming quadratic equations is incorrectly applying the sign for the coefficient of the 'x' term. Students frequently use x² + (sum of roots)x + (product of roots) = 0 instead of the correct form x² - (sum of roots)x + (product of roots) = 0. This error is compounded when the roots themselves are negative, leading to a miscalculation of the sum's sign before it's multiplied by the formula's inherent negative.
💭 Why This Happens:
  • Misremembering the Standard Form: Students often forget the crucial negative sign that precedes the sum of roots in the standard formula.
  • Confusion with General Form: Confusion can arise with the general quadratic form ax² + bx + c = 0, where 'b' is simply a coefficient, leading them to directly substitute the sum of roots as 'b' without considering the formula's specific negative.
  • Careless Substitution: Errors in algebraic manipulation when roots are negative can lead to incorrect calculation of the sum's sign.
✅ Correct Approach:
Always strictly adhere to the standard formula for a quadratic equation with roots α and β: x² - (α + β)x + (αβ) = 0. The negative sign before the sum of roots (α + β) is paramount. Calculate the sum of roots first, then substitute this value, ensuring the negative sign from the formula is applied.
📝 Examples:
❌ Wrong:
Let roots be α = -2 and β = 3.
Sum of roots (α + β) = -2 + 3 = 1
Product of roots (αβ) = (-2) * 3 = -6

Incorrect Application: Mistakenly using x² + (sum)x + (product) = 0
x² + (1)x + (-6) = 0
x² + x - 6 = 0 (Incorrect Equation)
✅ Correct:
Let roots be α = -2 and β = 3.
Sum of roots (α + β) = -2 + 3 = 1
Product of roots (αβ) = (-2) * 3 = -6

Correct Application: Using the formula x² - (sum)x + (product) = 0
x² - (1)x + (-6) = 0
x² - x - 6 = 0 (Correct Equation)
💡 Prevention Tips:
  • Memorize and Understand: Commit the formula x² - (sum of roots)x + (product of roots) = 0 to memory, focusing on the derivation from (x - α)(x - β) = 0.
  • Step-by-Step Calculation: Always calculate the sum (α + β) and product (αβ) separately and correctly, paying close attention to signs, especially with negative roots.
  • Careful Substitution: Substitute the calculated sum and product into the formula precisely. If the sum is negative (e.g., -5), the term becomes -(-5)x = +5x.
  • Self-Verification (JEE Advanced Tip): After forming the equation, quickly substitute one of the given roots back into the equation to verify if it satisfies the equation. For JEE Advanced, this quick check can save valuable marks.
JEE_Advanced
Important Sign Error

❌ Sign Error in Applying Sum of Roots Formula

Students frequently make sign errors when forming a quadratic equation from given roots, particularly with the term involving the sum of roots. They often forget the negative sign in the standard quadratic equation format.
✅ Correct Approach:
Always apply the correct standard form for a quadratic equation with roots α and β:
x2 - (Sum of Roots)x + (Product of Roots) = 0
Substitute the calculated sum and product carefully, paying close attention to the negative sign before the sum.
📝 Examples:
❌ Wrong:
Problem: Form the quadratic equation with roots 3 and -5.
Calculation: Sum of roots (α+β) = 3 + (-5) = -2, Product of roots (αβ) = 3 × (-5) = -15.
Incorrect Attempt: Using x2 + (α+β)x + αβ = 0
x2 + (-2)x + (-15) = 0
Incorrect Equation: x2 - 2x - 15 = 0
✅ Correct:
Problem: Form the quadratic equation with roots 3 and -5.
Calculation: Sum of roots (α+β) = 3 + (-5) = -2, Product of roots (αβ) = 3 × (-5) = -15.
Correct Application: Using x2 - (α+β)x + αβ = 0
x2 - (-2)x + (-15) = 0
Correct Equation: x2 + 2x - 15 = 0
💡 Prevention Tips:
  • Memorize the Formula: Commit x2 - (Sum of Roots)x + (Product of Roots) = 0 to memory.
  • Double-Check Signs: After calculating the sum and product, carefully substitute them into the formula and verify the signs.
  • Work Step-by-Step: Avoid mental calculations for substitution; write down each step to minimize errors.
  • Verify with Roots: For competitive exams like JEE, if time permits, quickly substitute one of the given roots back into your derived equation to see if it satisfies the equation.
JEE_Main
Important Unit Conversion

❌ Incorrect Signs in the General Form of Quadratic Equation

A very common error is to use incorrect signs when substituting the sum and product of roots into the general formula for forming a quadratic equation. Students often write x2 + (sum of roots)x + (product of roots) = 0 instead of the correct x2 - (sum of roots)x + (product of roots) = 0, or make errors with the sign of the product term.
💭 Why This Happens:
  • Lack of Formula Recall: Students might not have memorized the standard form x2 - (α+β)x + αβ = 0 correctly.
  • Misunderstanding Derivation: Forgetting that the formula originates from (x - α)(x - β) = 0, which upon expansion naturally yields the negative sign for the sum term.
  • Carelessness with Negative Numbers: When the sum or product of roots itself is negative, a double negative sign can be overlooked (e.g., -(-3) becoming -3 instead of +3).
  • Confusing with Coefficient Relations: While α+β = -b/a and αβ = c/a for ax2 + bx + c = 0, students sometimes mix these relations directly into the formation formula without accounting for the inherent negative sign.
✅ Correct Approach:
Always remember and apply the standard formula for forming a quadratic equation whose roots are α and β:
x2 - (Sum of Roots)x + (Product of Roots) = 0
Substitute the calculated values for the sum (α+β) and product (αβ) very carefully, paying close attention to their signs and the inherent negative sign in the formula before the sum term.
📝 Examples:
❌ Wrong:
Given roots are 3 and -5.
Sum of roots (α+β) = 3 + (-5) = -2
Product of roots (αβ) = 3 * (-5) = -15

Incorrect Attempt: Using x2 + (sum)x + (product) = 0
x2 + (-2)x + (-15) = 0
x2 - 2x - 15 = 0
(Here, the student incorrectly used '+' before the sum term. While the result is correct in this case because the sum was already negative, the conceptual mistake in applying the formula's sign convention is present.)
✅ Correct:
Given roots are 3 and -5.
Sum of roots (α+β) = 3 + (-5) = -2
Product of roots (αβ) = 3 * (-5) = -15

Using the correct formula x2 - (α + β)x + (αβ) = 0:
x2 - (-2)x + (-15) = 0
x2 + 2x - 15 = 0
The correct quadratic equation is x2 + 2x - 15 = 0.
💡 Prevention Tips:
  • Memorize the Exact Formula: Commit x2 - (sum of roots)x + (product of roots) = 0 to memory, especially the minus sign.
  • Understand the 'Why': Recall that if α and β are roots, then (x-α)(x-β) = 0 is the fundamental form. Expanding this will always yield x2 - (α+β)x + αβ = 0.
  • Substitute with Parentheses: Always use parentheses when substituting negative sums or products (e.g., -(-2) or +(-15)) to avoid sign errors.
  • Cross-Check in JEE Main: For MCQ questions, incorrect sign options are common distractors. Quickly verify by plugging one root back into your derived equation.
JEE_Main
Important Formula

❌ Sign Error in the Sum of Roots Term

Students often incorrectly form the quadratic equation as x² + (Sum of Roots)x + (Product of Roots) = 0. The crucial negative sign before the 'sum of roots' term in the correct formula, x² - (Sum of Roots)x + (Product of Roots) = 0, is frequently missed or misremembered.
💭 Why This Happens:
This error usually arises from incomplete or inaccurate memorization. Students often confuse it with the general form ax² + bx + c = 0, leading to direct substitution of the sum without the necessary negative sign.
✅ Correct Approach:
The correct formula for forming a quadratic equation with roots α and β is:
x² - (α + β)x + (αβ) = 0
Remember, the coefficient of 'x' is minus the sum of the roots, and the constant term is the product. This comes directly from expanding the factorized form (x - α)(x - β) = 0. For JEE Main, this fundamental understanding is critical.
📝 Examples:
❌ Wrong:
Given roots: 2 and 3.
Sum of roots: 2 + 3 = 5
Product of roots: 2 × 3 = 6
Incorrect equation formed: x² + 5x + 6 = 0. (Roots of this equation are -2 and -3, not 2 and 3).
✅ Correct:
Given roots: 2 and 3.
Sum of roots: 2 + 3 = 5
Product of roots: 2 × 3 = 6
Correct equation formed: x² - (5)x + (6) = 0, which simplifies to x² - 5x + 6 = 0. (Verify: For x² - 5x + 6 = 0, sum of roots = -(-5)/1 = 5, product of roots = 6/1 = 6).
💡 Prevention Tips:
  • Visualize Derivation: Always recall that the formula originates from (x - α)(x - β) = 0. Expanding this directly gives x² - (α + β)x + αβ = 0, solidifying the negative sign.
  • Cross-Check: After forming the equation, quickly find its roots or calculate the sum/product of roots from your derived equation. Compare these with the given values to confirm accuracy.
  • Memorize with Context: Instead of rote memorization, understand 'x² - (sum)x + (product) = 0' as a direct consequence of the root-factor relationship.
JEE_Main
Important Calculation

❌ Sign Errors in Substituting Sum of Roots

A frequent calculation error occurs when students correctly determine the sum (S) and product (P) of the given roots but make a sign mistake during their substitution into the standard quadratic equation form: x2 - Sx + P = 0. The most common pitfall is either omitting the negative sign before 'S' or misinterpreting it when 'S' itself is negative.
💭 Why This Happens:
This error primarily stems from:

  • Haste: Rushing through calculations, especially under exam pressure.
  • Formula Misremembering: Confusing x2 - Sx + P = 0 with x2 + Sx + P = 0.
  • Interaction of Signs: When the sum of roots (S) is negative, students might incorrectly substitute -S as - |S| instead of + |S|.
✅ Correct Approach:

To form a quadratic equation with given roots (let's say α and β):

  1. Calculate the Sum of Roots (S): S = α + β
  2. Calculate the Product of Roots (P): P = α × β
  3. Substitute Carefully: Always use the precise formula x2 - (Sum of roots)x + (Product of roots) = 0. Pay critical attention to the sign of 'S' and how it interacts with the negative sign in the formula. If 'S' is negative, -S will result in a positive term.
📝 Examples:
❌ Wrong:

Given roots: -2 and 5

1. Sum of roots (S) = -2 + 5 = 3

2. Product of roots (P) = (-2) × 5 = -10

Wrong Substitution: A common mistake is to write x2 + (3)x + (-10) = 0 leading to x2 + 3x - 10 = 0. Here, the negative sign before 'S' in the formula was ignored or incorrectly changed to positive.

✅ Correct:

Given roots: -2 and 5

1. Sum of roots (S) = -2 + 5 = 3

2. Product of roots (P) = (-2) × 5 = -10

3. Correct Substitution into x2 - Sx + P = 0:

x2 - (3)x + (-10) = 0

x2 - 3x - 10 = 0

💡 Prevention Tips:
  • Precise Formula Recall (JEE/CBSE): Always commit x2 - Sx + P = 0 to memory, specifically noting the crucial negative sign before the sum of roots term.
  • Sign Check (JEE): After calculating S and P, take a moment to double-check their signs and how they will interact with the formula's inherent negative sign during substitution. This is vital in JEE where small errors cost marks.
  • Quick Verification (JEE/CBSE): If time permits, especially in subjective board exams, you can quickly find the roots of the equation you formed (using the quadratic formula or factorization) to ensure they match the given roots.
JEE_Main
Important Conceptual

❌ Ignoring the General Constant Multiplier 'k'

Students frequently form a quadratic equation using the formula x2 - (sum of roots)x + (product of roots) = 0, but they often neglect the crucial arbitrary non-zero constant multiplier k. This oversight leads to forming only one specific equation, not the general form.
💭 Why This Happens:
This mistake stems from an over-reliance on simplified examples where the coefficient of x2 is implicitly taken as 1. Students often don't fully grasp that ax2 + bx + c = 0 and k(ax2 + bx + c) = 0 (for k ≠ 0) have the exact same roots, but represent different quadratic expressions (unless k=1).
✅ Correct Approach:
The correct general form of a quadratic equation with roots α and β is k[x2 - (α + β)x + αβ] = 0, where k is any non-zero real constant. In JEE Main, if the problem provides an additional condition (e.g., the coefficient of x2 is a specific value, or a point on the parabola is given), then k must be determined using that condition. If no such condition is given and only 'the' quadratic equation is asked, usually the simplest form (k=1 or k adjusted to make coefficients integers) is expected, but understanding the 'k' is conceptually vital.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3, a common incorrect approach is to state the equation as x2 - (2+3)x + (2×3) = 0, which simplifies to x2 - 5x + 6 = 0. This is only one of infinitely many such equations.
✅ Correct:
If the roots are 2 and 3, the general form of the quadratic equation is k(x2 - 5x + 6) = 0.
If the question additionally states, 'Find the quadratic equation whose roots are 2 and 3 and the coefficient of x2 is 4', then we set k=4, leading to 4(x2 - 5x + 6) = 0, or 4x2 - 20x + 24 = 0.
💡 Prevention Tips:
  • Always start with the general form: k[x2 - (Sum of Roots)x + (Product of Roots)] = 0.
  • Read carefully: Check for any additional constraints in the problem statement that might help determine the value of 'k'.
  • Conceptual understanding: Remember that multiplying an equation by a non-zero constant does not change its roots, but it changes the coefficients and hence the specific quadratic expression.
JEE_Main
Important Other

❌ Sign Error in the Sum of Roots Term

Students frequently correctly calculate the sum and product of the given roots but then incorrectly apply the sign when substituting the sum into the standard quadratic equation formation formula. Instead of using x² - (sum of roots)x + (product of roots) = 0, they often mistakenly write x² + (sum of roots)x + (product of roots) = 0. This leads to an incorrect coefficient for the 'x' term.
💭 Why This Happens:
  • Misremembering the Formula: The most common reason is a simple recall error, forgetting the negative sign that precedes the sum of roots term.
  • Confusion with Factor Form: While the factor form (x - α)(x - β) = 0 is correct, students might directly apply the sum/product values without understanding its expansion, leading to sign errors.
  • Lack of Derivational Understanding: Without understanding why the negative sign appears (from the expansion of (x-α)(x-β)), it's easier to make a mistake under exam pressure.
✅ Correct Approach:
The correct approach involves using the established formula for forming a quadratic equation with roots α and β consistently.

The standard formula is:

x² - (α + β)x + (αβ) = 0


Ensure that the sum of roots (α + β) is correctly calculated, and then its negative is taken for the coefficient of the 'x' term.
📝 Examples:
❌ Wrong:

Problem: Form the quadratic equation whose roots are 2 and 7.

Incorrect Solution:
Given roots: α = 2, β = 7
Sum of roots (α + β) = 2 + 7 = 9
Product of roots (αβ) = 2 × 7 = 14
Using the incorrect formula x² + (sum)x + (product) = 0:
x² + (9)x + (14) = 0
Result: x² + 9x + 14 = 0 (Incorrect)

✅ Correct:

Problem: Form the quadratic equation whose roots are 2 and 7.

Correct Solution:
Given roots: α = 2, β = 7
Sum of roots (α + β) = 2 + 7 = 9
Product of roots (αβ) = 2 × 7 = 14
Using the correct formula x² - (sum)x + (product) = 0:
x² - (9)x + (14) = 0
Result: x² - 9x + 14 = 0 (Correct)

💡 Prevention Tips:
  • Memorize and Understand: Commit the formula x² - (sum of roots)x + (product of roots) = 0 to memory, paying special attention to the minus sign. Understand its derivation from (x-α)(x-β)=0.
  • Double-Check Signs: Always explicitly write down the sum and product of roots. Then, when substituting into the formula, consciously check the sign of the 'x' term.
  • Verification: After forming the equation, quickly find its roots (by factorization or quadratic formula) to verify if they match the given roots. This acts as a robust self-check for both CBSE and JEE preparation.
CBSE_12th
Important Approximation

❌ Incorrect Application of Signs in the Quadratic Formula

Students frequently make sign errors when forming a quadratic equation using the sum and product of its roots. The most common mistake is using a positive sign before the sum of roots term, instead of the correct negative sign.
💭 Why This Happens:
This mistake primarily stems from:
  • A superficial memorization of the formula `x² + (sum)x + (product) = 0` instead of the correct `x² - (sum)x + (product) = 0`.
  • Lack of understanding of the formula's derivation from `(x - α)(x - β) = 0`.
  • Carelessness during substitution, especially when roots themselves are negative, leading to double negative errors or incorrect positive signs.
✅ Correct Approach:
The general form of a quadratic equation with roots α and β is given by:
x² - (α + β)x + αβ = 0.
Always remember the negative sign before the term containing the sum of roots. This is critical for both CBSE and JEE examinations.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3:
Sum of roots (α + β) = 2 + 3 = 5
Product of roots (αβ) = 2 × 3 = 6
Incorrect equation formed: x² + 5x + 6 = 0
(Here, the student incorrectly used a positive sign for the sum of roots term).
✅ Correct:
If the roots are 2 and 3:
Sum of roots (α + β) = 2 + 3 = 5
Product of roots (αβ) = 2 × 3 = 6
Correct equation formed: x² - (5)x + 6 = 0
which simplifies to x² - 5x + 6 = 0.

Another example, if roots are -2 and 3:
Sum of roots (α + β) = -2 + 3 = 1
Product of roots (αβ) = (-2) × 3 = -6
Correct equation formed: x² - (1)x + (-6) = 0
which simplifies to x² - x - 6 = 0.
💡 Prevention Tips:
  • Memorize Correctly: Firmly embed the formula x² - (Sum of Roots)x + (Product of Roots) = 0 into your memory.
  • Understand Derivation: Recall that the formula comes from `(x - α)(x - β) = 0` which expands to `x² - αx - βx + αβ = 0`, hence `x² - (α + β)x + αβ = 0`. This derivation reinforces the negative sign.
  • Double-Check Signs: Always pay extra attention to signs, especially when substituting negative values for roots.
  • JEE Callout: While the formula gives a basic quadratic equation, remember that `k(x² - (Sum)x + (Product)) = 0` (where `k` is any non-zero constant) also has the same roots. This is important if a specific leading coefficient is provided or required.
CBSE_12th
Important Sign Error

❌ Incorrect Sign Application for the Sum of Roots Term

Students frequently make sign errors when applying the formula for forming a quadratic equation, specifically in the term involving the sum of the roots. The standard form is x² - (sum of roots)x + (product of roots) = 0. A common mistake is to write x² + (sum of roots)x + (product of roots) = 0, or to incorrectly handle the sign when the sum of roots itself is negative, leading to an incorrect coefficient for the 'x' term.
💭 Why This Happens:
This error primarily stems from:
  • Rote Memorization: Students might remember the formula as having a '+' before the sum of roots term, or forget the crucial '-' sign.
  • Algebraic Oversight: When the sum of roots (α + β) is negative, for example, -5, students might incorrectly substitute it as '-5x' instead of '-(-5)x', which simplifies to '+5x'. They forget that the negative sign in the formula must be applied to the *entire* sum.
  • Rushing: Under exam pressure, students often overlook careful sign management.
✅ Correct Approach:
Always remember and consistently apply the correct formula for forming a quadratic equation with roots α and β:
x² - (α + β)x + αβ = 0.
When substituting the sum of roots (α + β), pay close attention to its calculated sign. If (α + β) is negative, the '-(α + β)' part will become positive.
📝 Examples:
❌ Wrong:
Problem: Form a quadratic equation whose roots are -2 and -3.
Incorrect Approach:
Sum of roots (α + β) = (-2) + (-3) = -5
Product of roots (αβ) = (-2) × (-3) = 6
Incorrectly applying the formula as x² + (sum)x + (product) = 0, or making a sign error in the '-(sum)' part:
x² + (-5)x + 6 = 0
x² - 5x + 6 = 0 (Incorrect 'x' term coefficient)
✅ Correct:
Problem: Form a quadratic equation whose roots are -2 and -3.
Correct Approach:
Sum of roots (α + β) = (-2) + (-3) = -5
Product of roots (αβ) = (-2) × (-3) = 6
Using the correct formula: x² - (α + β)x + αβ = 0
Substitute the values carefully:
x² - (-5)x + 6 = 0
x² + 5x + 6 = 0 (Correct equation)
💡 Prevention Tips:
  • Memorize Correctly: Ensure you remember the formula as x² - (sum of roots)x + (product of roots) = 0, paying specific attention to the minus sign before the sum term.
  • Calculate Sum and Product Separately: First, calculate the exact values of (α + β) and (αβ) with their correct signs.
  • Substitute with Parentheses: When substituting the sum of roots, especially if it's negative, use parentheses: x² - ( calculated_sum )x + ( calculated_product ) = 0. This helps avoid sign errors during simplification.
  • Double-Check: After forming the equation, quickly expand it back to verify the roots or compare the coefficients with Vieta's formulas.
CBSE_12th
Important Calculation

❌ Incorrect Sign Usage in Sum and Product of Roots

Students frequently make errors in sign conventions when substituting the sum (α + β) and product (αβ) of roots into the general quadratic equation formula: x² - (α + β)x + αβ = 0. This is particularly common when one or both roots are negative, leading to an incorrect sign for the 'x' term or the constant term.
💭 Why This Happens:
  • Carelessness with Formula: Students often forget the inherent negative sign in the formula for the 'x' term, i.e., -(sum of roots)x.
  • Direct Substitution Errors: Instead of first calculating the sum and product, they try to substitute roots directly, especially when dealing with negative numbers, leading to errors like x² + (α + β)x... instead of x² - (α + β)x....
  • Lack of Parentheses: Not using parentheses when substituting negative sums or products can cause calculation mistakes.
✅ Correct Approach:
Always follow these steps:
  • Step 1: Clearly identify the given roots, let them be α and β.
  • Step 2: Calculate the sum of roots: (α + β). Pay close attention to signs.
  • Step 3: Calculate the product of roots: (αβ). Pay close attention to signs.
  • Step 4: Substitute these calculated values (including their signs) into the formula: x² - (sum of roots)x + (product of roots) = 0.
📝 Examples:
❌ Wrong:

Given roots: α = 3, β = -5

Incorrect approach:

Sum = 3 + (-5) = -2

Product = (3)(-5) = -15

Equation: x² + (-2)x + (-15) = 0 (Incorrect: used + instead of - for the 'x' term)

Result: x² - 2x - 15 = 0

✅ Correct:

Given roots: α = 3, β = -5

Correct approach:

1. Calculate Sum: (α + β) = 3 + (-5) = -2

2. Calculate Product: (αβ) = (3)(-5) = -15

3. Use the formula: x² - (sum of roots)x + (product of roots) = 0

Substitute: x² - (-2)x + (-15) = 0

Simplify: x² + 2x - 15 = 0

💡 Prevention Tips:
  • CBSE & JEE Tip: Always write down the general formula (x² - (α + β)x + αβ = 0) before substituting. This reinforces the correct sign for the 'x' term.
  • Double Check: After calculating the sum and product, mentally or explicitly verify their signs before plugging them into the formula.
  • Warning: When the sum of roots is negative, the 'x' term in the equation will become positive (e.g., -(-2)x = +2x). Be extra vigilant here.
CBSE_12th
Important Formula

❌ Incorrect Sign in the Linear Term of the Quadratic Equation

A very common error is incorrectly applying the sign of the linear term ('x' term) when forming a quadratic equation from its roots. Students often use x² + (Sum of Roots)x + (Product of Roots) = 0 instead of the correct x² - (Sum of Roots)x + (Product of Roots) = 0.
💭 Why This Happens:
This mistake primarily arises from rote memorization without understanding the formula's derivation. The standard form of a quadratic equation is ax² + bx + c = 0, where the sum of roots (α + β) = -b/a. When 'a' is normalized to 1, the equation becomes x² + (b/a)x + (c/a) = 0, which means x² - (-b/a)x + (c/a) = 0. Confusing the direct substitution with the definition of 'b' (which is -a(sum of roots)) leads to this error. It's crucial to remember that the coefficient of 'x' directly corresponds to the negative of the sum of roots.
✅ Correct Approach:
Always use the fundamental formula for forming a quadratic equation when roots α and β are given:

x² - (α + β)x + (αβ) = 0

Here, (α + β) is the sum of the roots, and (αβ) is the product of the roots. The negative sign before the sum of roots is critical. For CBSE and JEE, this sign must be correctly applied to avoid incorrect equations.
📝 Examples:
❌ Wrong:
Given roots are 2 and 3.
Sum of roots (α + β) = 2 + 3 = 5.
Product of roots (αβ) = 2 * 3 = 6.
Wrong Equation: x² + 5x + 6 = 0. (The roots of this equation are -2 and -3).
✅ Correct:
Given roots are 2 and 3.
Sum of roots (α + β) = 2 + 3 = 5.
Product of roots (αβ) = 2 * 3 = 6.
Correct Equation: x² - 5x + 6 = 0. (Factoring gives (x-2)(x-3)=0, which yields roots 2 and 3).
💡 Prevention Tips:
  • Derive it once: Understand that the equation (x - α)(x - β) = 0 expands to x² - βx - αx + αβ = 0, which simplifies to x² - (α + β)x + αβ = 0. This reinforces the correct sign.
  • Check Your Answer: After forming the equation, quickly find its roots or substitute the given roots back into your equation to verify it.
  • Mind the Negatives: Be extra careful when the given roots themselves are negative, as this affects the sum and product, and subsequently, the signs in the final equation.
CBSE_12th
Important Unit Conversion

❌ Ignoring Implicit Unit Consistency in Roots (Application Context)

Students sometimes overlook that if roots are derived from quantities that inherently have units (even if not explicitly stated as 'units' alongside numerical values in simple formation problems, but implied by a broader problem context), and these quantities are expressed in different units, they must be converted to a consistent system before being used in Vieta's formulas. While rare for direct 'formation of quadratic equations with given roots' problems in CBSE which typically provide unitless numbers, this conceptual error can arise in more integrated application-based questions where roots represent physical magnitudes.
💭 Why This Happens:
  • Students often focus purely on the numerical values given, ignoring any implicit context of measurement or units that might influence the true magnitude.
  • Lack of practice with problems that integrate mathematical concepts with physics or real-world contexts, where unit consistency is paramount.
  • Assuming roots are always 'unitless' numbers, even when they might represent specific physical quantities like lengths, times, or masses in an underlying problem.
✅ Correct Approach:
  • Analyze the source or context of the roots: If roots represent physical quantities, ensure they are expressed in a consistent system of units before any calculations.
  • Perform necessary unit conversions: Convert all root values to a common, standardized unit (e.g., SI units) before calculating their sum and product.
  • Apply Vieta's formulas: Use the numerically consistent and correctly scaled root values in the standard formula: x² - (α + β)x + αβ = 0.
📝 Examples:
❌ Wrong:
Suppose a problem implicitly suggests roots are '2 meters' and '50 centimeters', and a student directly uses the numerical values '2' and '50' without unit conversion.
Given roots: α = 2 (from 2 m), β = 50 (from 50 cm)

Sum of roots (α+β) = 2 + 50 = 52

Product of roots (αβ) = 2 × 50 = 100

Quadratic equation: x² - 52x + 100 = 0
This approach is incorrect because the numerical values '2' and '50' are not consistent for mathematical operations unless their underlying units are the same.
✅ Correct:
Using the same roots, '2 meters' and '50 centimeters', but with correct unit conversion to ensure consistency.
Convert to a consistent unit (e.g., meters or centimeters).

Let's convert to meters:

α = 2 m

β = 50 cm = 0.5 m



Sum of roots (α+β) = 2 + 0.5 = 2.5

Product of roots (αβ) = 2 × 0.5 = 1



Quadratic equation: x² - 2.5x + 1 = 0

(To express with integer coefficients, multiply by 2: 2x² - 5x + 2 = 0)
💡 Prevention Tips:
  • Thorough Problem Reading: Always read the problem statement carefully, especially for application-based questions, to identify if the roots represent physical quantities with implied units.
  • Standardize Units Early: Make it a standard practice to convert all related quantities to a single, consistent system of units (e.g., SI units) at the very beginning of problem-solving.
  • Contextual Awareness: Understand that mathematical problems can abstract real-world scenarios. Attention to details like unit consistency, even if not explicitly stated in the 'given roots' part, is crucial for accurate results.
CBSE_12th
Critical Conceptual

❌ Ignoring the General Form `k(x^2 - (Sum of Roots)x + Product of Roots) = 0`

A common critical conceptual mistake is assuming that if α and β are the roots of a quadratic equation, the equation is *always* `x^2 - (α + β)x + αβ = 0`. Students often overlook the crucial leading coefficient 'k' (where `k` is any non-zero real number), which represents the family of all quadratic equations with those specific roots. This oversight becomes critical when additional conditions are provided, such as a specific leading coefficient, or the equation passing through a given point.
💭 Why This Happens:
This mistake typically arises from over-simplification during initial learning, where the standard form `x^2 - (sum)x + (product) = 0` is memorized without a complete understanding of its general derivation or the implications of scaling. Students often confuse 'a' quadratic equation with 'the' specific quadratic equation that satisfies all given conditions.
✅ Correct Approach:
Always start with the general form of the quadratic equation: k(x^2 - (Sum of Roots)x + Product of Roots) = 0, where `k` is a non-zero constant.
  • First, calculate the sum (S = α + β) and product (P = αβ) of the given roots.
  • Substitute S and P into the general form: `k(x^2 - Sx + P) = 0`.
  • If additional conditions are given (e.g., the equation's leading coefficient is specified, or it passes through a particular point), use these conditions to determine the unique value of 'k'. If no additional conditions are given, any non-zero 'k' provides a valid quadratic equation.
📝 Examples:
❌ Wrong:
Problem: Form *a* quadratic equation whose roots are 2 and 3.
Student's Approach:
Sum of roots (S) = 2 + 3 = 5.
Product of roots (P) = 2 × 3 = 6.
Equation is `x^2 - 5x + 6 = 0`.
Mistake: While `x^2 - 5x + 6 = 0` is *a* correct equation, this approach often leads students to believe it's the *only* one, failing to generalize to `k(x^2 - 5x + 6) = 0` when further conditions are required.
✅ Correct:
Problem: Form the quadratic equation whose roots are 2 and 3, and for which the coefficient of `x^2` is 2.
Correct Approach:
1. Calculate Sum (S) = 2 + 3 = 5.
2. Calculate Product (P) = 2 × 3 = 6.
3. Write the general form: `k(x^2 - Sx + P) = 0`.
4. Substitute S and P: `k(x^2 - 5x + 6) = 0`.
5. Expand the equation: `kx^2 - 5kx + 6k = 0`.
6. Given that the coefficient of `x^2` is 2, we have `k = 2`.
7. Substitute `k=2` back into the equation: `2(x^2 - 5x + 6) = 0`, which simplifies to `2x^2 - 10x + 12 = 0`.
💡 Prevention Tips:
  • Conceptual Clarity: Understand that `x^2 - Sx + P = 0` represents *one* such equation (with k=1), but `k(x^2 - Sx + P) = 0` represents the entire family.
  • JEE Specific: In JEE Main, questions often involve conditions that require determining 'k'. Always begin with the `k` factor to avoid missing crucial information.
  • Practice: Solve problems where 'k' needs to be found using additional conditions (e.g., passing through a point, or specific coefficients for terms other than `x^2`).
JEE_Main
Critical Formula

❌ Incorrect Sign in Sum of Roots Term (x² - Sx + P = 0)

A very common and critical error students make when forming a quadratic equation from given roots (α and β) is misremembering the sign before the 'sum of roots' term. The standard formula is x² - (α + β)x + (αβ) = 0, but students often incorrectly write it as x² + (α + β)x + (αβ) = 0, especially if the sum of roots itself is negative.
💭 Why This Happens:
This mistake primarily stems from rote memorization of the formula without understanding its derivation from the factored form (x - α)(x - β) = 0. Students might also confuse it with other polynomial relations or simply forget the specific negative sign, leading to incorrect quadratic equations. A negative sum of roots often exacerbates this error, as students instinctively think 'plus a negative' and end up with 'plus' when it should remain 'minus' followed by the negative sum.
✅ Correct Approach:
Always remember the correct general form for a quadratic equation with roots α and β:
x² - (Sum of Roots)x + (Product of Roots) = 0
Where,
  • Sum of Roots (S) = α + β
  • Product of Roots (P) = αβ
Substitute the calculated sum and product carefully, paying close attention to the sign of the sum of roots. The negative sign outside the parenthesis for 'Sum of Roots' is crucial and constant.
📝 Examples:
❌ Wrong:
If roots are 2 and -5:
Sum (S) = 2 + (-5) = -3
Product (P) = 2 * (-5) = -10
Incorrect Equation: x² + (-3)x + (-10) = 0
Which simplifies to: x² - 3x - 10 = 0 (This looks correct, but the initial substitution showed the error in thinking +S)

A clearer error: If roots are -2 and -5:
Sum (S) = -2 + (-5) = -7
Product (P) = (-2) * (-5) = 10
Incorrect Equation Attempt: x² + (-7)x + 10 = 0
Resulting in: x² - 7x + 10 = 0. This would be the correct final equation by chance, if the student just replaced (S) directly. However, the conceptual error is trying to apply 'x² + Sx + P = 0' instead of 'x² - Sx + P = 0'. A student might write x² + (-7)x + 10 = 0 which then simplifies correctly. The common error happens when students write x² + (7)x + 10 = 0 thinking the formula is always positive for 'S'.
✅ Correct:
If roots are -2 and -5:
Sum (S) = (-2) + (-5) = -7
Product (P) = (-2) * (-5) = 10
Correct Equation: x² - (Sum of Roots)x + (Product of Roots) = 0
x² - (-7)x + (10) = 0
Simplifies to: x² + 7x + 10 = 0
💡 Prevention Tips:
  • Understand the Derivation: Remember the quadratic equation comes from (x - α)(x - β) = 0. Expanding this gives x² - (α + β)x + αβ = 0.
  • Double-Check the Sign: Always explicitly write out the 'minus' sign before the sum of roots, even if the sum itself is negative, e.g., x² - (negative sum)x + product = 0.
  • Practice with Varied Roots: Work through examples with positive, negative, and zero roots to solidify the application of the formula.
  • Recite the Formula: Verbally repeat 'x squared MINUS sum of roots x PLUS product of roots equals zero' to embed the correct signs.
CBSE_12th
Critical Other

❌ Sign Error in the Coefficient of 'x'

Students frequently make a critical error by using a positive sign instead of a negative sign for the sum of roots when forming a quadratic equation. The standard formula is x2 - (Sum of Roots)x + (Product of Roots) = 0, but students often incorrectly write x2 + (Sum of Roots)x + (Product of Roots) = 0.
💭 Why This Happens:
This error primarily stems from a lack of understanding of the derivation of the formula or confusion with Vieta's formulas. In Vieta's formulas, for ax2 + bx + c = 0, the sum of roots is -b/a. Students might mistakenly substitute this sum directly into the general form without accounting for the inherent negative sign in the equation formation formula.
✅ Correct Approach:
Always recall the fundamental property: if α and β are the roots of a quadratic equation, then the equation can be formed as (x - α)(x - β) = 0. Expanding this expression yields x2 - (α + β)x + αβ = 0. This clearly shows that the coefficient of 'x' is always the negative of the sum of the roots. This foundational understanding is crucial for both CBSE and JEE examinations.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3:
Incorrect thought process: Sum = 2 + 3 = 5, Product = 2 × 3 = 6.
Incorrect Equation: x2 + 5x + 6 = 0
✅ Correct:
If the roots are 2 and 3:
Correct thought process: Sum = 2 + 3 = 5, Product = 2 × 3 = 6.
Correct Equation: x2 - (5)x + 6 = 0 ⇒ x2 - 5x + 6 = 0
💡 Prevention Tips:
  • Understand the Derivation: Consistently remember that the formula originates from (x - α)(x - β) = 0.
  • Write Down the Formula: Before substituting values, always explicitly write down the general form: x2 - (Sum)x + (Product) = 0.
  • Double-Check Signs: Be extremely vigilant with signs, especially when the roots themselves are negative or involve complex/irrational numbers (e.g., if one root is 2-&sqrt;3, the other is 2+&sqrt;3 for real coefficients, and their sum is 4).
  • Verify: After forming the equation, quickly substitute one of the given roots into the equation to confirm it satisfies the equation.
CBSE_12th
Critical Sign Error

❌ Critical Sign Error: Incorrect Sum of Roots Substitution

Students frequently make a critical sign error when substituting the sum of roots into the standard quadratic equation formula, x2 - (α + β)x + αβ = 0. This mistake is particularly common and severe when one or both roots are negative, leading to an incorrect sign for the middle term (coefficient of x).
💭 Why This Happens:
  • Forgetting the Formula's Negative Sign: The most common reason is overlooking the intrinsic negative sign in x2 - (Sum)x + Product = 0.
  • Arithmetic Errors with Signed Numbers: Incorrectly adding or subtracting negative numbers during the calculation of the sum of roots.
  • Rushing Calculations: A lack of careful attention to detail, especially under exam pressure, leads to slips.
  • Confusing Algebraic Sum with Absolute Sum: Some students might incorrectly take the absolute value of the sum, ignoring the actual algebraic signs of the roots.
✅ Correct Approach:

To correctly form the quadratic equation:

  1. Identify the Roots: Let the given roots be α and β.
  2. Calculate the Sum of Roots: Determine α + β. Pay meticulous attention to the signs of α and β.
  3. Calculate the Product of Roots: Determine α × β.
  4. Substitute into the Formula: Place the calculated sum and product into the formula: x2 - (α + β)x + αβ = 0. Always remember the minus sign before (α + β)x.
📝 Examples:
❌ Wrong:

Given roots: 2 and -5

1. Sum of roots (Incorrect Calculation): Student might incorrectly calculate 2 + 5 = 7 (ignoring the negative sign), or more commonly, calculate 2 + (-5) = -3, but then write the equation as x2 + (-3)x + (-10) = 0 which simplifies to x2 - 3x - 10 = 0.

Error: The crucial mistake is writing + (-3)x instead of - (-3)x, failing to apply the formula's inherent negative sign correctly.

✅ Correct:

Given roots: 2 and -5

1. Sum of roots (α + β): 2 + (-5) = -3

2. Product of roots (αβ): 2 × (-5) = -10

3. Substitute into the formula: x2 - (α + β)x + αβ = 0

x2 - (-3)x + (-10) = 0

Correct Equation: x2 + 3x - 10 = 0

For CBSE 12th, this correct application is vital, as a single sign error can lead to a completely wrong answer.

💡 Prevention Tips:
  • Strictly Memorize the Formula: Reiterate x2 - (Sum)x + (Product) = 0. Focus on the negative sign before the sum.
  • Use Parentheses for Substitution: Always enclose the sum and product in parentheses when substituting, especially if they are negative. E.g., x2 - (-3)x + (-10) = 0.
  • Double-Check Each Sign: After writing the final equation, mentally or physically re-check the signs of each term against the formula and your calculated sum/product.
  • Practice with Varied Roots: Work through problems with positive, negative, and fractional roots to build confidence and accuracy in sign handling.
CBSE_12th
Critical Unit Conversion

❌ Sign Error in the Coefficient of 'x' Term

Students frequently make a critical sign error when forming a quadratic equation from given roots. Instead of using the correct standard form x2 - (sum of roots)x + (product of roots) = 0, they mistakenly use x2 + (sum of roots)x + (product of roots) = 0.
💭 Why This Happens:
This error often stems from misremembering the standard formula. While the product of roots term (constant term) always retains its natural sign, the coefficient of the 'x' term is the negative of the sum of the roots. Students might conflate this with simply adding the sum of roots without the crucial negative sign, leading to an incorrect quadratic equation.
✅ Correct Approach:
Always remember and apply the standard form for a quadratic equation with roots α and β:
x2 - (α + β)x + αβ = 0
Here,
  • (α + β) is the sum of the roots.
  • αβ is the product of the roots.
CBSE/JEE Tip: This formula is fundamental. Even if given a specific leading coefficient 'a', the structure remains a[x2 - (α + β)x + αβ] = 0.
📝 Examples:
❌ Wrong:
Given roots: α = 3, β = -5
Sum of roots (α + β): 3 + (-5) = -2
Product of roots (αβ): 3 × (-5) = -15
Incorrect formation: x2 + (-2)x + (-15) = 0
x2 - 2x - 15 = 0 ✗ (Incorrect!)
✅ Correct:
Given roots: α = 3, β = -5
Sum of roots (α + β): 3 + (-5) = -2
Product of roots (αβ): 3 × (-5) = -15
Correct formation using x2 - (α + β)x + αβ = 0:
x2 - (-2)x + (-15) = 0
x2 + 2x - 15 = 0 ✓ (Correct!)
💡 Prevention Tips:
  • Memorize Correctly: Consciously remember the negative sign: x2 - (sum)x + (product) = 0.
  • Double-Check: After forming the equation, quickly find its roots using factorization or the quadratic formula to verify if they match the given roots.
  • Practice: Solve multiple problems, especially with negative and fractional roots, to solidify the correct formula application.
CBSE_12th
Critical Conceptual

❌ Sign Error in the Linear Term (Sum of Roots)

A frequent and critical conceptual error is incorrectly applying the negative sign associated with the sum of roots in the standard quadratic equation formula. Students often use x² + (Sum of Roots)x + (Product of Roots) = 0 instead of the correct x² - (Sum of Roots)x + (Product of Roots) = 0, especially when the sum of roots itself is negative.
💭 Why This Happens:
This mistake stems from a misunderstanding of the derivation of the formula or simply misremembering the exact form. The root-factor form (x-α)(x-β) = 0 expands to x² - (α+β)x + αβ = 0. The negative sign before (α+β) is intrinsic. If students just calculate the sum (S) and product (P) and then try to fit them into a loosely remembered `x² ± Sx + P = 0` form, they often get the sign of the linear term wrong.
✅ Correct Approach:
Always strictly adhere to the standard formula for forming a quadratic equation when given its roots α and β:
x² - (α + β)x + (αβ) = 0
Where,
  • Sum of Roots (S) = α + β
  • Product of Roots (P) = αβ
The equation then becomes: x² - Sx + P = 0. Pay extreme attention when substituting a negative sum of roots into the formula.
📝 Examples:
❌ Wrong:
Problem: Form the quadratic equation with roots α = -2 and β = -3.
Incorrect Steps:
1. Calculate Sum of Roots (S) = (-2) + (-3) = -5.
2. Calculate Product of Roots (P) = (-2) × (-3) = 6.
3. Student incorrectly applies the formula as x² + Sx + P = 0.
4. Incorrect Equation: x² + (-5)x + 6 = 0 which simplifies to x² - 5x + 6 = 0.
✅ Correct:
Problem: Form the quadratic equation with roots α = -2 and β = -3.
Correct Steps:
1. Calculate Sum of Roots (S) = (-2) + (-3) = -5.
2. Calculate Product of Roots (P) = (-2) × (-3) = 6.
3. Apply the correct formula: x² - Sx + P = 0.
4. Correct Equation: x² - (-5)x + 6 = 0 which simplifies to x² + 5x + 6 = 0.
💡 Prevention Tips:
  • Memorize and Understand: Firmly engrave the formula x² - (Sum of Roots)x + (Product of Roots) = 0 in your mind. Understand its derivation from (x-α)(x-β)=0.
  • Step-by-Step Calculation: Always calculate the sum (S) and product (P) of the roots separately first, paying close attention to the signs.
  • Careful Substitution: When substituting S and P into the formula, explicitly write out the negative sign from the formula. For example, if S = -5, write -(-5) to ensure the correct positive sign for the linear term.
  • Self-Check: After forming the equation, you can quickly substitute one of the roots back into it to verify if the equation holds true.
CBSE_12th
Critical Calculation

❌ Sign and Arithmetic Errors in Calculating Sum and Product of Roots

Students frequently make critical calculation errors when determining the sum (α + β) and product (αβ) of given roots, especially when dealing with negative numbers, fractions, or a combination thereof. These errors directly lead to incorrect coefficients in the quadratic equation x² - (α + β)x + αβ = 0.
💭 Why This Happens:
  • Carelessness with Signs: Forgetting to carry negative signs during addition or multiplication, or incorrect application of sign rules (e.g., negative times negative is positive).
  • Fractional Arithmetic Errors: Mistakes in finding common denominators, adding/subtracting fractions, or multiplying fractions.
  • Rushing: Students often rush calculations under exam pressure, leading to oversight of basic arithmetic principles.
  • Lack of Double-Check: Not verifying the computed sum and product before forming the equation.
✅ Correct Approach:
Always calculate the sum and product of roots meticulously, step by step. Write down intermediate steps clearly, paying close attention to signs and fractional operations. A systematic approach helps minimize errors.
The standard form for an equation with roots α and β is:
x² - (Sum of Roots)x + (Product of Roots) = 0
x² - (α + β)x + (αβ) = 0
📝 Examples:
❌ Wrong:
Problem: Form a quadratic equation with roots -3/4 and 2/5.
Student's Incorrect Calculation:
Sum of roots (α + β) = -3/4 + 2/5 = (-15 - 8)/20 = -23/20 (Error: incorrect addition of -8)
Product of roots (αβ) = (-3/4) * (2/5) = 6/20 = 3/10 (Error: sign omitted)
Equation formed: x² - (-23/20)x + (3/10) = 0
                     20x² + 23x + 6 = 0 (Incorrect final equation)
✅ Correct:
Problem: Form a quadratic equation with roots -3/4 and 2/5.
Correct Calculation:
Sum of roots (α + β) = -3/4 + 2/5 = (-15 + 8)/20 = -7/20
Product of roots (αβ) = (-3/4) * (2/5) = -6/20 = -3/10
Substitute into the formula x² - (α + β)x + αβ = 0:
x² - (-7/20)x + (-3/10) = 0
x² + (7/20)x - (3/10) = 0
To clear fractions (often required for JEE Main options), multiply by the LCM of denominators (20):
20(x²) + 20(7/20)x - 20(3/10) = 0
20x² + 7x - 6 = 0 (Correct final equation)
💡 Prevention Tips:
  • Slow Down and Focus: Take your time with arithmetic, especially when negative numbers or fractions are involved.
  • Write All Steps: Do not attempt mental calculations for sum and product, especially if the roots are complex. Write each step down.
  • Use Parentheses: When substituting negative values or expressions, use parentheses to avoid sign errors, e.g., x² - (-7/20)x.
  • Double-Check Signs: After calculating the sum and product, consciously review the signs. For multiplication, remember:
    • + × + = +
    • - × - = +
    • + × - = -
  • Verify with Options (JEE Specific): If multiple-choice options are given, quickly check if your calculated coefficients match any option. This can sometimes flag a calculation error early.
JEE_Main
Critical Formula

❌ Confusing the Sign of the Sum of Roots in the Quadratic Equation Formula

A common and critical error is misremembering the standard formula for forming a quadratic equation when its roots, say α and β, are given. Students frequently write the equation as x² + (α + β)x + αβ = 0 instead of the correct x² - (α + β)x + αβ = 0. This leads to an incorrect coefficient for the 'x' term, rendering the entire equation (and subsequent calculations) wrong.
💭 Why This Happens:
  • Rote Memorization: Students often memorize the formula without understanding its derivation from (x - α)(x - β) = 0.
  • Carelessness: Under exam pressure, a small but critical sign can be easily overlooked.
  • Confusion: Sometimes, this sign confusion stems from vaguely remembering other algebraic identities or quadratic properties where signs might differ.
✅ Correct Approach:
Always remember and verify the standard formula for a quadratic equation with roots α and β:
x² - (Sum of Roots)x + (Product of Roots) = 0
or, more explicitly,
x² - (α + β)x + αβ = 0.
The negative sign before the sum of roots is crucial and directly comes from the expansion of (x - α)(x - β) = 0.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3:
Sum of roots (S) = 2 + 3 = 5
Product of roots (P) = 2 × 3 = 6

Incorrect approach: Forming the equation as x² + Sx + P = 0
Result: x² + 5x + 6 = 0
✅ Correct:
If the roots are 2 and 3:
Sum of roots (S) = 2 + 3 = 5
Product of roots (P) = 2 × 3 = 6

Correct approach: Using the formula x² - Sx + P = 0
Result: x² - 5x + 6 = 0
💡 Prevention Tips:
  • Understand Derivation: Spend a minute to derive the formula from (x - α)(x - β) = 0 to internalize why the 'x' term has a negative sign.
  • Highlight the Sign: Consciously focus on the negative sign for the sum of roots part of the formula.
  • Verification (JEE Main Tip): After forming the equation, quickly substitute one of the given roots back into your derived equation to check if it satisfies the equation. This simple step can catch major errors.
JEE_Main
Critical Unit Conversion

❌ Failing to Unify/Simplify Root Expressions Before Calculation (Analogous to Unit Conversion Error)

A common and critical error in JEE Main is when students directly substitute complex or unsimplified expressions for roots (e.g., radical expressions with unrationalized denominators, complex fractions) into the sum (α+β) and product (αβ) formulas. This is fundamentally similar to making unit conversion errors in physics problems; if the 'units' (the numerical forms of the roots) are not consistent or simplified to a common 'standard', calculations become erroneous and unnecessarily complicated.
💭 Why This Happens:
This mistake typically arises from rushing, a lack of methodical problem-solving, or not recognizing that algebraic expressions for roots need to be 'standardized' or 'converted' to their simplest numerical form first. Students often mechanically apply the quadratic equation formula x² - (α+β)x + αβ = 0 without ensuring that α and β are in their most manageable and consistent numerical representations, leading to complex intermediate steps and calculation blunders.
✅ Correct Approach:
The correct approach is to always simplify each given root expression to its simplest numerical form before performing any calculations for their sum and product. This often involves rationalizing denominators, simplifying radical terms, or evaluating complex fractions. Ensure both roots are in a fully reduced and consistent numerical representation, much like converting all measurements to SI units before calculation.
📝 Examples:
❌ Wrong:
Given roots: α = (√3 + 1) / 2 and β = 1 / (√3 - 1)
Student's mistake: Directly calculating Sum (α+β) and Product (αβ) with β in its unsimplified form.
Sum (α+β) = (√3 + 1)/2 + 1/(√3 - 1)
Product (αβ) = ((√3 + 1)/2) * (1/(√3 - 1))
This leads to cumbersome algebra and a high probability of errors.
✅ Correct:
Given roots: α = (√3 + 1) / 2 and β = 1 / (√3 - 1)
Correct approach:
1. Simplify β first:
β = 1 / (√3 - 1) = (1 * (√3 + 1)) / ((√3 - 1) * (√3 + 1)) = (√3 + 1) / ((√3)² - 1²) = (√3 + 1) / (3 - 1) = (√3 + 1) / 2
2. Now, both roots are in a simplified and consistent form: α = (√3 + 1) / 2 and β = (√3 + 1) / 2
3. Calculate sum: α + β = ((√3 + 1) / 2) + ((√3 + 1) / 2) = (2√3 + 2) / 2 = √3 + 1
4. Calculate product: αβ = ((√3 + 1) / 2) * ((√3 + 1) / 2) = (√3 + 1)² / 4 = (3 + 2√3 + 1) / 4 = (4 + 2√3) / 4 = (2 + √3) / 2
5. Form the quadratic equation: x² - (√3 + 1)x + ((2 + √3) / 2) = 0
2x² - 2(√3 + 1)x + (2 + √3) = 0 (Multiplying by 2 to clear the fraction).
💡 Prevention Tips:
  • JEE Tip: Always make a habit of scanning given root expressions for any simplification opportunities (e.g., rationalizing denominators, simplifying radicals, evaluating complex numbers or fractions) before proceeding to calculate their sum and product.
  • Treat the simplification step as a mandatory 'pre-processing' or 'unit conversion' of the roots into a standard, simplified numerical form.
  • Practice problems where roots are intentionally given in unsimplified forms to build this crucial analytical skill.
  • Double-check the simplified form of each root before substituting them into the sum and product formulas to avoid 'cascading' errors.
JEE_Main
Critical Other

❌ Neglecting the General Form with a Leading Coefficient 'k'

Students often assume that a quadratic equation with given roots $alpha$ and $eta$ is uniquely represented by $x^2 - (alpha + eta)x + alphaeta = 0$. They frequently miss the crucial leading constant 'k', i.e., k(x2 - (alpha + eta)x + alphaeta) = 0. This oversight becomes critical when additional conditions, such as the equation passing through a specific point or having a particular coefficient, are provided. This is a common error in JEE Advanced context.
💭 Why This Happens:
This error stems from an incomplete understanding of how a quadratic equation is formed. While $x^2 - ( ext{sum})x + ext{product} = 0$ provides a valid quadratic, it's just one of an infinite family of equations sharing the same roots, differing only by a non-zero scalar multiple 'k'. The standard general form $ax^2 + bx + c = 0$ implies that 'a' (which is 'k' here) can be any non-zero real number. Students often implicitly assume k=1.
✅ Correct Approach:
Always form the quadratic equation in its most general form: k(x2 - (Sum of Roots)x + Product of Roots) = 0, where 'k' is a non-zero constant. If no additional information is provided, setting k=1 is usually sufficient. However, if conditions like "the equation passes through a point (p, q)" or "the coefficient of x2 is 'A'" are given, then 'k' must be determined using these conditions.
(JEE Advanced Tip: Problems often include such conditions to test this very understanding.)
📝 Examples:
❌ Wrong:
  • Problem: Form a quadratic equation with roots 2 and 3, which passes through the point (1, 6).
  • Wrong Approach:
    Sum of Roots (S) = 2+3 = 5
    Product of Roots (P) = 2*3 = 6
    Equation: $x^2 - 5x + 6 = 0$.
    Substituting (1, 6) into the equation: $1^2 - 5(1) + 6 = 1 - 5 + 6 = 2$.
    Since $2
    eq 6$, this equation does not pass through the point (1, 6). The student failed to account for 'k'.
✅ Correct:
  • Correct Approach:
    Roots are 2 and 3. So, S = 5 and P = 6.
    General Equation: $k(x^2 - 5x + 6) = 0$.
    The equation passes through (1, 6). Substitute x=1 and the equation's value as 6:
    $k(1^2 - 5(1) + 6) = 6$
    $k(1 - 5 + 6) = 6$
    $k(2) = 6 implies k = 3$.
    Therefore, the correct equation is 3(x2 - 5x + 6) = 0, which simplifies to 3x2 - 15x + 18 = 0.
💡 Prevention Tips:
  • Always include 'k': When forming a quadratic equation from given roots, begin with the general form $k(x^2 - Sx + P) = 0$.
  • Read carefully: Always check if any extra conditions are given (e.g., passes through a point, specific coefficient of $x^2$, etc.) that require determining the value of 'k'.
  • Conceptual clarity: Understand that a quadratic equation is determined by its roots and an arbitrary non-zero scaling factor 'k'. For CBSE, k=1 is often assumed if not specified, but for JEE, 'k' is frequently tested.
JEE_Advanced
Critical Approximation

❌ Sign Error in the Linear Term and Neglecting the General Constant Factor 'k'

Students frequently make two critical errors when forming quadratic equations from given roots:
1. Incorrect sign for the coefficient of 'x': Instead of using - (Sum of Roots)x, they mistakenly use + (Sum of Roots)x. This fundamental sign error leads to a completely different quadratic equation.
2. Neglecting the general constant factor 'k': While x² - (Sum)x + (Product) = 0 is the monic form, any equation k(x² - (Sum)x + (Product)) = 0 where k ≠ 0 also has the same roots. In JEE Advanced, problems might require determining 'k' based on additional conditions (e.g., specific leading coefficient, or the equation passing through a given point). Ignoring 'k' (implicitly assuming k=1) without justification can lead to an incomplete or incorrect answer.
💭 Why This Happens:
  • Haste and lack of attention to detail: Under exam pressure, a crucial sign is often overlooked.
  • Rote memorization: Memorizing 'S' (sum) and 'P' (product) without a precise understanding of the quadratic formula's structure x² - Sx + P = 0.
  • Incomplete understanding of general form: Forgetting that ax² + bx + c = 0 is a general form and that k represents the leading coefficient 'a' (or a scalar multiple).
  • Approximation in thought process: Students might *approximately* recall the formula, leading to conceptual errors like sign flips or assuming the monic form without considering the generality of 'k'.
✅ Correct Approach:
  1. Precisely calculate the sum (S) and product (P) of the given roots. Pay meticulous attention to signs, fractions, and complex numbers.
  2. Always use the standard formula: k [x² - (Sum of Roots)x + (Product of Roots)] = 0.
  3. For CBSE and basic JEE questions: If not specified, 'k' is usually taken as 1, forming the monic equation (e.g., x² - Sx + P = 0).
  4. For JEE Advanced: If the question asks for an equation with specific properties (e.g., coefficient of x² is 2, or the equation passes through a point), then 'k' must be explicitly determined. If no such condition is given, the simplest form (k=1, or clearing fractions) is acceptable, but be aware of the 'k' factor.
📝 Examples:
❌ Wrong:
Given roots are -3 and 5.
Sum S = -3 + 5 = 2.
Product P = (-3)(5) = -15.
Incorrect equation formed: x² + 2x - 15 = 0 (Common sign error for the 'x' term).

Another mistake: If asked for an equation with a leading coefficient of 2, students might incorrectly write x² - 2x - 15 = 0 and then multiply by 2 to get 2x² - 4x - 30 = 0, not realizing that if they had first written k(x² - 2x - 15) = 0 and then put k=2, it would have been more robust.
✅ Correct:
Given roots are -3 and 5.
Sum S = -3 + 5 = 2.
Product P = (-3)(5) = -15.
The correct general quadratic equation is: k [x² - Sx + P] = 0.
Substituting S and P: k [x² - (2)x + (-15)] = 0.
Simplifying: k (x² - 2x - 15) = 0.
If 'k' is not specified, the monic equation (k=1) is: x² - 2x - 15 = 0.
If the question specifies a leading coefficient of 2, then k=2:
2 (x² - 2x - 15) = 0, which simplifies to 2x² - 4x - 30 = 0.
💡 Prevention Tips:
  • Strictly memorize the formula: Reiterate x² - (Sum of Roots)x + (Product of Roots) = 0. Pay explicit attention to the negative sign before the sum term.
  • Always verify: Mentally (or on scratch paper) check the roots of your formed equation using Vieta's formulas (-b/a and c/a) to confirm they match the given roots.
  • Understand 'k' contextually: For JEE Advanced, always consider the general factor 'k'. Determine if the question implies a specific 'k' or if the monic form (k=1) is sufficient. Read the question carefully for any additional conditions.
  • Practice diverse problems: Work through examples with integer, fractional, irrational, and complex roots to ensure the formula and its application are deeply ingrained, preventing approximation errors in understanding.
JEE_Advanced
Critical Sign Error

❌ Sign Errors in Forming Quadratic Equations

A common and critically severe mistake students make when forming a quadratic equation from given roots (say, α and β) is getting the sign wrong for the coefficient of the 'x' term. Instead of using the correct standard form, x2 - (α + β)x + αβ = 0, students often mistakenly use x2 + (α + β)x + αβ = 0 or other incorrect sign combinations.

💭 Why This Happens:
  • Misremembering the Standard Form: Students often forget the explicit minus sign before the sum of roots in the standard equation x2 - (Sum)x + (Product) = 0.
  • Carelessness with Negative Roots: When one or both roots are negative, the sum of roots (α + β) might itself be negative. Forgetting to apply the 'minus of a negative' (e.g., - (-3) = +3) leads to errors.
  • Direct Substitution without Pre-calculation: Rushing to substitute roots directly into the formula without first calculating the explicit sum (S) and product (P) values separately.
✅ Correct Approach:

To prevent sign errors, always follow these steps:

  1. Calculate the Sum of Roots (S): Add the given roots carefully, paying attention to their signs (S = α + β).
  2. Calculate the Product of Roots (P): Multiply the given roots carefully, paying attention to their signs (P = αβ).
  3. Substitute into Standard Form: Use the universally correct form:
    x2 - Sx + P = 0
    The minus sign before 'S' is crucial. If 'S' itself is negative, the term -Sx will become positive (e.g., - (-3)x = +3x).
📝 Examples:
❌ Wrong:

Problem: Form a quadratic equation whose roots are -2 and 3.

Common Wrong Approach:
Sum of roots (S) = -2 + 3 = 1
Product of roots (P) = (-2)(3) = -6
Mistakenly applies as x2 + Sx + P = 0
Equation formed: x2 + 1x + (-6) = 0 which simplifies to x2 + x - 6 = 0. This is incorrect.

✅ Correct:

Problem: Form a quadratic equation whose roots are -2 and 3.

Correct Approach:
Given roots: α = -2, β = 3
1. Calculate Sum (S): S = α + β = -2 + 3 = 1
2. Calculate Product (P): P = αβ = (-2)(3) = -6
3. Substitute into the correct standard form: x2 - Sx + P = 0
x2 - (1)x + (-6) = 0
Simplifying yields: x2 - x - 6 = 0. This is the correct quadratic equation.

💡 Prevention Tips:
  • Memorize the Standard Form: Engrain x2 - (Sum)x + (Product) = 0 firmly in memory.
  • Separate Calculations: Always calculate S and P as distinct values before plugging them into the equation. This isolates potential sign errors to a single step.
  • Double-Check Signs: Be extra vigilant when one or both roots are negative. If the sum 'S' itself is negative, remember that -S will result in a positive coefficient for 'x'.
  • Practice with Variety: Work through problems with roots that are both positive, both negative, and one positive/one negative.
JEE_Advanced
Critical Unit Conversion

❌ <span style='color: red;'>Ignoring Unit Consistency for Roots Representing Physical Quantities</span>

Students often overlook the units associated with given roots, especially when they represent physical measurements. If the roots are provided in inconsistent units (e.g., one in meters and another in centimeters), direct calculation of the sum or product of roots without proper unit conversion will lead to incorrect numerical coefficients for the quadratic equation. This is a subtle trap in JEE Advanced.
💭 Why This Happens:
  • Lack of meticulous reading of problem statements, a common challenge in JEE Advanced where nuances are critical.
  • Assuming roots are always dimensionless pure numbers without considering the context.
  • Focusing solely on numerical values rather than the complete information, including units.
  • Underestimating the importance of unit consistency in mathematical problems with physical relevance.
✅ Correct Approach:
  • Always scrutinize the problem statement to identify if the roots represent physical quantities and are accompanied by units.
  • If units are present and inconsistent, convert all roots to a common standard unit *before* calculating their numerical sum (S) and product (P).
  • Form the quadratic equation using the standard formula: x² - (Sum of Roots)x + (Product of Roots) = 0, where 'x' represents the numerical value of the root in the chosen consistent unit.
  • For JEE Advanced, such problems test conceptual clarity and attention to detail, making unit consistency a critical step.
📝 Examples:
❌ Wrong:
Given roots: α = 2 m and β = 50 cm.
Incorrect Student Calculation:
Sum (S) = 2 + 50 = 52
Product (P) = 2 * 50 = 100
Incorrect Equation: x² - 52x + 100 = 0 (This is wrong because the units were not made consistent before summing/multiplying, leading to incorrect numerical coefficients).
✅ Correct:
Given roots: α = 2 m and β = 50 cm.
  1. The problem typically expects the quadratic equation with dimensionless coefficients, where the variable 'x' represents the numerical value of the root in a consistent unit. Let's choose meters as the consistent unit.
  2. Convert β = 50 cm = 0.5 m.
  3. Now, the numerical values of the roots (in meters) are α' = 2 and β' = 0.5.
  4. Sum of numerical values (S) = 2 + 0.5 = 2.5.
  5. Product of numerical values (P) = 2 * 0.5 = 1.0.
  6. Correct Equation: x² - 2.5x + 1.0 = 0.

JEE Advanced Note: While direct formation of equations with dimensioned coefficients is rare in standard JEE problems, understanding unit consistency for extracting the numerical values of roots is crucial. This scenario highlights a critical attention-to-detail error that can arise if roots are presented with inconsistent units in a physics-contextualized math problem.

💡 Prevention Tips:
  • Read Carefully: Always check if the roots are merely numbers or represent quantities with units.
  • Standardize Units: If units are involved, ensure they are consistent across all roots *before* performing any arithmetic operations to derive sum and product.
  • Contextual Awareness: Consider the overall context of the problem in JEE Advanced; subtle clues often point to specific requirements like unit consistency for numerical calculations.
  • Double-Check: Before finalizing the equation, quickly re-verify that the roots' numerical values used for S and P were derived from consistent units.
JEE_Advanced
Critical Formula

❌ Sign Error in the Linear Term (Sum of Roots)

A frequent critical error in forming a quadratic equation from given roots (α and β) is incorrectly applying the sign for the linear term (coefficient of x). Students often use x2 + (Sum of Roots)x + (Product of Roots) = 0 instead of the correct form: x2 - (Sum of Roots)x + (Product of Roots) = 0.
💭 Why This Happens:
This mistake primarily stems from a lack of understanding the derivation of the formula. Students may rotely memorize the general form ax2 + bx + c = 0 where -b/a is the sum of roots, and mistakenly apply a positive 'b' directly. The crucial negative sign in x2 - (Sum of Roots)x + (Product of Roots) = 0 directly originates from expanding the factored form (x - α)(x - β) = 0.
✅ Correct Approach:
The fundamental formula for a quadratic equation with roots α and β is: x2 - (α + β)x + αβ = 0. Here, (α + β) represents the sum of the roots, and αβ represents the product of the roots. Always ensure the coefficient of the 'x' term has a negative sign before the sum of roots.
📝 Examples:
❌ Wrong:
If the roots are 2 and 3, a common incorrect formation would be:
  • Sum of roots (α + β) = 2 + 3 = 5
  • Product of roots (αβ) = 2 * 3 = 6
  • Incorrect equation: x2 + 5x + 6 = 0
✅ Correct:
For the same roots, 2 and 3:
  • Sum of roots (α + β) = 2 + 3 = 5
  • Product of roots (αβ) = 2 * 3 = 6
  • Correct equation: x2 - 5x + 6 = 0
💡 Prevention Tips:
  • Understand, don't just memorize: Always remember the expansion of (x - α)(x - β) = 0 explicitly leads to x2 - (α + β)x + αβ = 0.
  • Double-Check Signs: When substituting the sum of roots, particularly if the roots are negative or have varying signs, be extremely careful with the overall sign of the linear term.
  • JEE Advanced Note: A single sign error can lead to a completely incorrect quadratic equation, rendering all subsequent calculations wrong. Precision is paramount.
JEE_Advanced
Critical Calculation

❌ Incorrect Sign Convention in Sum/Product Calculation or Formula Application

Students frequently make critical sign errors when calculating the sum (α + β) or product (αβ) of given roots, especially with negative numbers or when roots involve irrational terms. A common mistake also involves incorrectly applying the general quadratic equation formula, x2 - (sum of roots)x + (product of roots) = 0, by using a '+' sign for the 'sum of roots' term instead of '-'.
💭 Why This Happens:
  • Haste and lack of attention to detail during calculations.
  • Weak foundational arithmetic skills, particularly with negative numbers.
  • Confusion or misremembering the standard formula for forming quadratic equations.
  • Overlooking the implications of the negative sign in x2 - (α + β)x + αβ = 0.
✅ Correct Approach:
Always meticulously calculate the sum and product of the given roots, paying extreme attention to their signs. Then, strictly apply the standard formula for forming a quadratic equation: ax2 + bx + c = 0, where b = -(α + β) and c = αβ (assuming a=1). This translates to x2 - (Sum of Roots)x + (Product of Roots) = 0. Remember the negative sign before the sum of roots is crucial.
📝 Examples:
❌ Wrong:
Given roots: α = -2, β = 3
Student's calculation:
Sum (α + β) = -2 + 3 = 1 (Correct sum)
Product (αβ) = -2 * 3 = 6 (Incorrect sign for product)
Equation formed: x2 - (1)x + 6 = 0 => x2 - x + 6 = 0 (Wrong final equation due to product sign error).

Another common wrong application:
Sum (α + β) = 1, Product (αβ) = -6
Student's formula application: x2 + (Sum)x + (Product) = 0
Equation formed: x2 + (1)x + (-6) = 0 => x2 + x - 6 = 0 (Wrong final equation due to formula sign error).
✅ Correct:
Given roots: α = -2, β = 3
Correct calculation:
Sum of roots (α + β) = -2 + 3 = 1
Product of roots (αβ) = -2 * 3 = -6
Applying the correct formula: x2 - (α + β)x + αβ = 0
x2 - (1)x + (-6) = 0
x2 - x - 6 = 0
This is the correct quadratic equation.
💡 Prevention Tips:
  • Write down each step: Clearly note down the roots, then sum, then product, and finally the equation.
  • Double-check signs: Pay extra attention to negative numbers during addition and multiplication.
  • Memorize the formula: Rehearse x2 - (Sum of Roots)x + (Product of Roots) = 0 until it's second nature. The negative sign for the linear term is critical.
  • Verify with mental substitution: If time permits, quickly check if the original roots satisfy the formed equation.
JEE_Advanced
Critical Conceptual

❌ Ignoring the Leading Coefficient 'k' and Sign Errors

Students frequently form the quadratic equation as x^2 - (sum of roots)x + (product of roots) = 0 and overlook the arbitrary non-zero constant k. The correct general form should be k[x^2 - (sum of roots)x + (product of roots)] = 0. Additionally, frequent sign errors occur in the middle term.
💭 Why This Happens:
This mistake stems from over-reliance on the basic x^2 - Sx + P = 0 form without fully grasping that multiplying a polynomial by a non-zero constant does not change its roots. Carelessness with algebraic signs also contributes to errors.
✅ Correct Approach:
  1. Calculate Sum (S) and Product (P): For given roots α and β, S = α + β and P = αβ.
  2. Apply General Form: The general quadratic equation is k[x² - Sx + P] = 0, where k is any non-zero real constant.
  3. Determine 'k': Use any additional conditions provided in the problem (e.g., the value of a specific coefficient, or a point the parabola passes through) to find the unique value of k.
  4. Sign Convention: Crucially, the 'x' term always has a minus sign in the x² - Sx + P = 0 formulation, i.e., - (Sum of roots)x.
📝 Examples:
❌ Wrong:
Problem: Form a quadratic equation with roots 2 and 3, and a leading coefficient of 2.
Student's Mistake:
Sum (S) = 2 + 3 = 5
Product (P) = 2 * 3 = 6
Equation formed: x² - 5x + 6 = 0.
Here, the student incorrectly assumes k=1, ignoring the given leading coefficient of 2. If the problem then asks for the coefficient of x, they'd incorrectly state -5 instead of -10.
✅ Correct:
Problem: Form a quadratic equation with roots 2 and 3, and a leading coefficient of 2.
Correct Approach:
  • Sum of roots (S) = 2 + 3 = 5
  • Product of roots (P) = 2 * 3 = 6
  • Start with the general form: k(x² - Sx + P) = 0
  • Substitute S and P: k(x² - 5x + 6) = 0
  • Given the leading coefficient is 2. The leading term is kx². So, we must have k = 2.
  • The correct quadratic equation is 2(x² - 5x + 6) = 0, which expands to 2x² - 10x + 12 = 0.
💡 Prevention Tips:
  • Always start with the general form k[x² - (α + β)x + αβ] = 0. This 'k' is often the key differentiator in JEE Advanced problems.
  • Double-check the sign for the 'x' term; it's always -(Sum of roots)x.
  • Utilize all given conditions (e.g., specific coefficient, a point the curve passes through) to accurately determine k.
JEE_Advanced
Critical Sign Error

❌ Sign Errors in Forming Quadratic Equations

Students often make critical sign errors when substituting the sum and product of roots into the standard quadratic equation form. The most common mistake is incorrectly placing the negative sign before the sum of roots term, leading to an incorrect quadratic equation. This is particularly prevalent when the calculated sum of roots itself is negative, causing confusion with double negatives.
💭 Why This Happens:
  • Rote Memorization without Understanding: Students might partially recall the formula x2 - (sum)x + (product) = 0 but forget the specific sign convention.
  • Carelessness: Rushing calculations during the exam can lead to overlooking the crucial negative sign associated with the 'sum of roots' term.
  • Confusion with Negative Roots: If the sum of roots (α+β) is, for example, -5, students might incorrectly write x2 - 5x + ... instead of x2 - (-5)x + ....
  • Lack of Verification: Not taking a moment to cross-check the signs after forming the equation.
✅ Correct Approach:
The fundamental principle for forming a quadratic equation with given roots α and β is to use the formula:
x2 - (α + β)x + (αβ) = 0.
Always follow these steps:
  1. Calculate the Sum (α+β): Carefully add the given roots.
  2. Calculate the Product (αβ): Carefully multiply the given roots.
  3. Substitute Carefully: Substitute the *exact calculated values* for (α+β) and (αβ) into the formula, paying close attention to the leading negative sign for the sum of roots term.
JEE Tip: This is a very basic but high-frequency error. A single sign error can render the entire answer incorrect, even if the magnitudes are right.
📝 Examples:
❌ Wrong:

Given Roots: α = -4, β = 2

Step 1: Sum (α+β) = -4 + 2 = -2

Step 2: Product (αβ) = (-4) × (2) = -8

Wrong Application: Substituting the sum incorrectly, e.g., x2 - 2x + (-8) = 0 which simplifies to x2 - 2x - 8 = 0. Here, the student used - (positive sum)x instead of - (negative sum)x.

✅ Correct:

Given Roots: α = -4, β = 2

Step 1: Sum (α+β) = -4 + 2 = -2

Step 2: Product (αβ) = (-4) × (2) = -8

Correct Application (using x2 - (sum)x + (product) = 0):
x2 - (-2)x + (-8) = 0
x2 + 2x - 8 = 0

💡 Prevention Tips:
  • Master the Formula: Ensure the formula x2 - (α+β)x + αβ = 0 is ingrained, especially the minus sign before the sum of roots.
  • Parentheses for Substitution: Always use parentheses when substituting the calculated sum and product, especially if they are negative. Example: x2 - (-2)x + (-8) = 0.
  • Quick Verification: Mentally (or on scratch paper) check the roots of the equation you formed using the quadratic formula, if time permits.
  • Derive if Doubtful: If you ever forget the sign, quickly derive it from (x-α)(x-β) = 0, which expands to x2 - αx - βx + αβ = 0 or x2 - (α+β)x + αβ = 0.
JEE_Main
Critical Approximation

❌ Incorrect Approximation of Irrational/Complex Roots

A critical mistake in JEE Main is approximating irrational numbers (e.g., √2, √3) or complex numbers when calculating the sum (S) and product (P) of given roots. Instead of using their exact algebraic forms, students sometimes convert them to decimal approximations. This leads to an approximate quadratic equation, which is almost always incorrect for JEE Main where exact answers are expected.
💭 Why This Happens:
This error stems from a lack of precision in algebraic manipulation and a misconception that all numerical values should be converted to decimal forms. Over-reliance on calculators for intermediate steps can also introduce rounding errors. Furthermore, not fully understanding the simplification offered by conjugate pairs (e.g., a + √b and a - √b, or a + ib and a - ib) contributes to this problem.
✅ Correct Approach:
Always use the exact algebraic forms of the roots (including irrational or complex parts) to calculate the sum (S) and product (P). For roots α and β, the quadratic equation is x² - Sx + P = 0, where S = α + β and P = αβ. When roots are in conjugate pairs, their sum and product simplify significantly, resulting in rational or real coefficients and avoiding square roots or imaginary parts in the final equation.
📝 Examples:
❌ Wrong:
Given roots are (2 + √5) and (2 - √5).
Wrong Approximation: Approximate √5 ≈ 2.236.
  • S ≈ (2 + 2.236) + (2 - 2.236) = 4.236 - 0.236 = 4.000
  • P ≈ (2 + 2.236)(2 - 2.236) = 4.236 * (-0.236) ≈ -1.000
Equation: x² - 4x - 1 = 0 (Incorrect due to rounding)
✅ Correct:
Given roots are (2 + √5) and (2 - √5).
Correct Exact Method: Use exact forms.
  • S = (2 + √5) + (2 - √5) = 2 + 2 + √5 - √5 = 4
  • P = (2 + √5)(2 - √5) = 2² - (√5)² = 4 - 5 = -1
Equation: x² - 4x - 1 = 0 (This is the exact correct equation)
💡 Prevention Tips:
  • JEE Tip: Always work with exact values of irrational numbers and complex numbers. Only approximate if the question specifically asks for an approximate value, which is rare for equation formation.
  • Recognize Conjugate Pairs: Understand that if one root is a + √b, the other root is often a - √b (or a + ib and a - ib), making sum and product calculations straightforward and exact.
  • Avoid Calculators for Intermediate Steps: Do not rely on calculators for calculations that might introduce rounding errors with irrational or complex numbers.
  • Practice Exact Algebra: Strengthen your skills in manipulating expressions involving square roots and complex numbers precisely.
JEE_Main

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Formation of quadratic equations with given roots

Subject: Mathematics
Unit: 2 - COMPLEX NUMBERS AND QUADRATIC EQUATIONS
Sub-unit: 2.2 - Quadratic Equations
Complexity: Mid
Syllabus: JEE_Main

Content Completeness: 66.7%

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📝 CBSE Problems: 12
🎯 JEE Problems: 18
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⚠️ Mistakes: 58
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