How to Solve Quadratic Equations (Nature of Roots, Sum/Product, Transformation, Location of Roots) – IIT JEE Guide

How to Solve Quadratic Equations (Nature of Roots, Sum/Product, Transformation, Location of Roots) – IIT JEE Guide

Introduction

Mastering quadratic equations doesn’t just get you 10-15 marks in IIT JEE—it unlocks entire problems in algebra, coordinate geometry, and calculus. One wrong assumption about roots? You lose 5+ marks in a single question. Today, you’ll learn exactly how to analyze roots, transform equations, and spot examiner traps—so you never lose marks again.

What You Need To Know First

1. Standard form of a quadratic equation: ( ax^2 + bx + c = 0 ) (where ( a \neq 0 )).
2. Discriminant (( D )): ( D = b^2 - 4ac ). Determines the nature of roots.
3. Sum and product of roots: For roots ( \alpha ) and ( \beta ), ( \alpha + \beta = -\frac{b}{a} ), ( \alpha \beta = \frac{c}{a} ).

Key Vocabulary

Formulas To Know

Step-by-Step Method

1. Nature of Roots (Using Discriminant)

Step 1: Write the quadratic in standard form: ( ax^2 + bx + c = 0 ). Step 2: Calculate ( D = b^2 - 4ac ). Step 3: Check the sign of ( D ): - ( D > 0 ): Two distinct real roots. - ( D = 0 ): One real root (repeated). - ( D < 0 ): Two complex roots.

Example: Find the nature of roots of ( 2x^2 - 4x + 3 = 0 ). Step 1: ( a = 2, b = -4, c = 3 ). Step 2: ( D = (-4)^2 - 4(2)(3) = 16 - 24 = -8 ). Step 3: ( D < 0 ), so roots are complex.

2. Sum and Product of Roots

Step 1: For ( ax^2 + bx + c = 0 ), sum ( \alpha + \beta = -\frac{b}{a} ). Step 2: Product ( \alpha \beta = \frac{c}{a} ).

Example: Find sum and product of roots of ( x^2 - 6x + 8 = 0 ). Step 1: Sum ( = -\frac{-6}{1} = 6 ). Step 2: Product ( = \frac{8}{1} = 8 ).

3. Forming a Quadratic Equation with Given Roots

Step 1: If roots are ( \alpha, \beta ), equation is ( x^2 - (\alpha + \beta)x + \alpha \beta = 0 ). Step 2: Substitute sum and product.

Example: Form a quadratic equation with roots 3 and -2. Step 1: Sum ( = 3 + (-2) = 1 ), Product ( = 3 \times (-2) = -6 ). Step 2: Equation: ( x^2 - (1)x + (-6) = 0 ) → ( x^2 - x - 6 = 0 ).

4. Transformation of Roots

Case 1: Roots Multiplied by a Constant (( k ))

Step 1: Original roots ( \alpha, \beta ). Step 2: New roots ( k\alpha, k\beta ). Step 3: New equation: ( x^2 - k(\alpha + \beta)x + k^2 \alpha \beta = 0 ).

Example: Original equation: ( x^2 - 5x + 6 = 0 ) (roots 2, 3). Find equation with roots doubled. Step 1: Sum ( = 5 ), Product ( = 6 ). Step 2: New sum ( = 2 \times 5 = 10 ), New product ( = 4 \times 6 = 24 ). Step 3: New equation: ( x^2 - 10x + 24 = 0 ).

Case 2: Roots Shifted by a Constant (( c ))

Step 1: Original roots ( \alpha, \beta ). Step 2: New roots ( \alpha + c, \beta + c ). Step 3: Substitute ( x ) with ( x - c ) in original equation.

Example: Original equation: ( x^2 - 5x + 6 = 0 ). Find equation with roots increased by 1. Step 1: Replace ( x ) with ( x - 1 ). Step 2: ( (x - 1)^2 - 5(x - 1) + 6 = 0 ). Step 3: Simplify: ( x^2 - 2x + 1 - 5x + 5 + 6 = 0 ) → ( x^2 - 7x + 12 = 0 ).

5. Location of Roots (Between Two Numbers)

Condition 1: Both roots between ( p ) and ( q ). - ( D \geq 0 ) (real roots). - ( f(p) > 0 ) and ( f(q) > 0 ) (same sign at endpoints). - ( p < -\frac{b}{2a} < q ) (vertex between ( p ) and ( q )).

Condition 2: One root between ( p ) and ( q ), other outside. - ( f(p) \times f(q) < 0 ) (sign change).

Example: Check if ( x^2 - 4x + 3 = 0 ) has roots between 1 and 3. Step 1: ( D = 16 - 12 = 4 > 0 ) (real roots). Step 2: ( f(1) = 1 - 4 + 3 = 0 ), ( f(3) = 9 - 12 + 3 = 0 ). Step 3: Roots are exactly at 1 and 3 (not strictly between).

Worked Examples

Example 1 – Basic (Nature of Roots)

Question: Find the nature of roots of ( 3x^2 + 2x - 1 = 0 ). Solution:
1. ( a = 3, b = 2, c = -1 ).
2. ( D = 2^2 - 4(3)(-1) = 4 + 12 = 16 ).
3. ( D > 0 ), so two distinct real roots.

What we did and why: We used the discriminant to check if roots are real or complex. Since ( D > 0 ), roots are real and different.

Example 2 – Medium (Transformation)

Question: If roots of ( x^2 - 5x + 6 = 0 ) are ( \alpha, \beta ), find the equation with roots ( 2\alpha + 1, 2\beta + 1 ). Solution:
1. Original sum ( \alpha + \beta = 5 ), product ( \alpha \beta = 6 ).
2. New roots: ( 2\alpha + 1, 2\beta + 1 ).
3. New sum ( = 2(\alpha + \beta) + 2 = 2(5) + 2 = 12 ).
4. New product ( = (2\alpha + 1)(2\beta + 1) = 4\alpha \beta + 2(\alpha + \beta) + 1 = 4(6) + 2(5) + 1 = 24 + 10 + 1 = 35 ).
5. New equation: ( x^2 - 12x + 35 = 0 ).

What we did and why: We transformed the roots by scaling and shifting, then used sum and product to form the new equation.

Example 3 – Exam-Style (Location of Roots)

Question: For ( f(x) = x^2 - 2x - 3 ), check if one root lies between 1 and 4. Solution:
1. ( f(1) = 1 - 2 - 3 = -4 ).
2. ( f(4) = 16 - 8 - 3 = 5 ).
3. ( f(1) \times f(4) = (-4)(5) = -20 < 0 ).
4. Since sign changes, one root lies between 1 and 4.

What we did and why: We used the Intermediate Value Theorem (sign change) to confirm a root exists in the interval.

Common Mistakes

Exam Traps

1-Minute Recap

Listen up—this is your last-minute checklist:
1. Nature of roots? Calculate ( D = b^2 - 4ac ). ( D > 0 ): two real roots. ( D = 0 ): one real root. ( D < 0 ): complex roots.
2. Sum and product? Sum ( = -\frac{b}{a} ), product ( = \frac{c}{a} ). Use these to form new equations.
3. Transforming roots? Scale: multiply sum by ( k ), product by ( k^2 ). Shift: replace ( x ) with ( x - c ).
4. Location of roots? Check ( f(p) \times f(q) < 0 ) for one root in ( (p, q) ). For both roots, check ( D \geq 0 ), ( f(p) > 0 ), ( f(q) > 0 ), and vertex in ( (p, q) ).
5. Exam traps? Watch for disguised quadratics, non-integer shifts, and boundary roots.

You’ve got this. Now go solve those quadratic problems like a pro.