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Basics Solutions Discriminant Vieta's Formation Graph Inequalities Applications

Quadratic Equations JEE Main & Advanced 2025-26

Master one of the most important chapters with complete notes on Discriminant, Roots Analysis, Vieta's Formulas, Graph & Applications. Includes 200+ solved problems and all JEE shortcuts.

📚 8 Complete Sections
✍️ 200+ Solved Examples
🎯 Quick Formulas
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1

Quadratic Equation Fundamentals

A quadratic equation is a second-degree polynomial equation in one variable. It's one of the most fundamental concepts in algebra and appears extensively in JEE Main and Advanced.

1.1 Definition and Standard Form

Standard Form of Quadratic Equation

\[ax^2 + bx + c = 0\]

Where:

  • a, b, c are real constants (coefficients)
  • a ≠ 0 (must be non-zero, otherwise it becomes linear)
  • x is the variable (unknown)
  • a is the coefficient of x² (leading coefficient)
  • b is the coefficient of x (linear coefficient)
  • c is the constant term (free term)
✅ Valid Quadratic Equations

x² + 5x + 6 = 0

a=1, b=5, c=6

2x² - 7x + 3 = 0

a=2, b=-7, c=3

x² - 16 = 0

a=1, b=0, c=-16

-3x² + 4x = 0

a=-3, b=4, c=0

❌ NOT Quadratic Equations

5x + 7 = 0

Linear (degree 1)

x³ + 2x² + 1 = 0

Cubic (degree 3)

0·x² + 5x + 2 = 0

a = 0, becomes linear

√x + 5 = 0

Not polynomial form

1.2 Important Terminology

Term Definition Example
Roots/Zeros Values of x that satisfy the equation (make it zero) x² - 5x + 6 = 0 has roots 2, 3
Solutions Same as roots (interchangeable term) x = 2 and x = 3 are solutions
Degree Highest power of variable (always 2 for quadratic) Degree = 2
Coefficients Constants a, b, c in ax² + bx + c = 0 In 3x² + 5x - 2 = 0: a=3, b=5, c=-2
Monic Equation Quadratic equation with a = 1 x² + 3x + 2 = 0
Pure Quadratic When b = 0 (no linear term) 4x² - 25 = 0

1.3 Types of Quadratic Equations

Complete Quadratic

All three terms present (a, b, c all non-zero)

ax² + bx + c = 0

Example:

2x² + 5x + 3 = 0

Pure Quadratic

Only x² and constant term (b = 0)

ax² + c = 0

Example:

3x² - 27 = 0

x² - 16 = 0

Incomplete Quadratic

Missing constant term (c = 0)

ax² + bx = 0

Example:

x² + 5x = 0

2x² - 8x = 0

📝 Solved Example 1

Question: Convert the following equations to standard form and identify coefficients a, b, c:
(a) (x - 3)(x + 5) = 0
(b) (2x + 1)² = 9
(c) x(x - 4) = 21

Solution:

(a) (x - 3)(x + 5) = 0

Expanding: x² + 5x - 3x - 15 = 0

x² + 2x - 15 = 0

a = 1, b = 2, c = -15

(b) (2x + 1)² = 9

Expanding: 4x² + 4x + 1 = 9

4x² + 4x + 1 - 9 = 0

4x² + 4x - 8 = 0

a = 4, b = 4, c = -8

(c) x(x - 4) = 21

Expanding: x² - 4x = 21

x² - 4x - 21 = 0

a = 1, b = -4, c = -21

💡 Key Points to Remember

  • Always write in standard form: ax² + bx + c = 0 before solving
  • Sign matters: Pay attention to signs of b and c coefficients
  • a cannot be zero: If a = 0, it's not quadratic
  • Maximum 2 roots: A quadratic equation has at most 2 solutions
  • Coefficient identification: Practice identifying a, b, c quickly for JEE
2

Methods to Solve Quadratic Equations

There are four main methods to solve quadratic equations. Choosing the right method based on the equation type is crucial for solving JEE problems quickly.

2.1 Method 1: Factorization

Factorization Method

Express ax² + bx + c as a product of two linear factors and apply zero product property.

Steps:

  1. Write equation in standard form: ax² + bx + c = 0
  2. Find two numbers that multiply to give a×c and add to give b
  3. Split the middle term using these numbers
  4. Factor by grouping
  5. Apply zero product property: If AB = 0, then A = 0 or B = 0

📝 Solved Example 2 - Factorization

Question: Solve by factorization: x² + 5x + 6 = 0

Solution:

Step 1: Identify coefficients

a = 1, b = 5, c = 6

Step 2: Find two numbers that:

• Multiply to give a × c = 1 × 6 = 6

• Add to give b = 5

Numbers are: 2 and 3 (2 × 3 = 6, 2 + 3 = 5)

Step 3: Split the middle term

x² + 2x + 3x + 6 = 0

Step 4: Factor by grouping

x(x + 2) + 3(x + 2) = 0

(x + 2)(x + 3) = 0

Step 5: Apply zero product property

x + 2 = 0 OR x + 3 = 0

x = -2 OR x = -3

Answer: x = -2, -3

2.2 Method 2: Completing the Square

Completing the Square Method

Convert the equation into perfect square form: (x + p)² = q

For equation: ax² + bx + c = 0

  1. Divide throughout by 'a' if a ≠ 1
  2. Move constant to right side
  3. Add (b/2a)² to both sides
  4. Write left side as perfect square
  5. Take square root and solve
\[x^2 + \frac{b}{a}x = -\frac{c}{a}\] \[x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}\] \[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\]

2.3 Method 3: Quadratic Formula (Sridharacharya's Formula)

⭐ MOST IMPORTANT - Quadratic Formula

For equation ax² + bx + c = 0, the roots are given by:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Understanding the formula:

  • ± means two roots: one with + and one with -
  • b² - 4ac is called the discriminant (D or Δ)
  • This formula works for ALL quadratic equations
  • Derived from completing the square method

The two roots are:

α = (-b + √D)/(2a)

β = (-b - √D)/(2a)

📝 Solved Example 3 - Quadratic Formula

Question: Solve using quadratic formula: 2x² + 7x + 3 = 0

Solution:

Step 1: Identify coefficients

a = 2, b = 7, c = 3

Step 2: Calculate discriminant

D = b² - 4ac

D = (7)² - 4(2)(3)

D = 49 - 24 = 25

Step 3: Apply quadratic formula

\[x = \frac{-7 \pm \sqrt{25}}{2(2)} = \frac{-7 \pm 5}{4}\]

Step 4: Find both roots

First root (taking +):

x₁ = (-7 + 5)/4 = -2/4 = -1/2

Second root (taking -):

x₂ = (-7 - 5)/4 = -12/4 = -3

Answer: x = -1/2, -3

2.4 Method 4: Special Cases (Quick Methods)

Pure Quadratic (b = 0)

Form: ax² + c = 0

ax² = -c

x² = -c/a

x = ±√(-c/a)

Example: x² - 25 = 0

x² = 25

x = ±5

Missing Constant (c = 0)

Form: ax² + bx = 0

x(ax + b) = 0

x = 0 or ax + b = 0

x = 0 or x = -b/a

Example: 3x² - 12x = 0

x(3x - 12) = 0

x = 0 or x = 4

💡 Which Method to Use? (JEE Strategy)

Situation Best Method Time
Easy to factor (small integers) Factorization Fastest
b = 0 (pure quadratic) Direct method Fastest
c = 0 (missing constant) Taking x common Fastest
Decimal/fraction coefficients Quadratic Formula Medium
Complex looking equation Quadratic Formula Medium
Nature of roots questions Discriminant only Very Fast
3

Discriminant and Nature of Roots

The discriminant is the most important concept for determining root characteristics WITHOUT actually solving the equation. This is heavily tested in JEE.

3.1 The Discriminant (D or Δ)

⭐ Discriminant Formula

For quadratic equation ax² + bx + c = 0:

\[D = b^2 - 4ac\]

Why it's called "Discriminant":

It discriminates (distinguishes) between different types of roots based on its value.

3.2 Nature of Roots Based on Discriminant

Discriminant Value Nature of Roots Type Example
D > 0 Two distinct real roots Real & Unequal x² - 5x + 6 = 0
D = 25-24 = 1 > 0
D > 0 & perfect square Two distinct rational roots Rational & Unequal x² - 5x + 6 = 0
D = 1 (perfect square)
D = 0 Two equal real roots (repeated root) Real & Equal x² - 6x + 9 = 0
D = 36-36 = 0
D < 0 No real roots (complex conjugate) Imaginary/Complex x² + 2x + 5 = 0
D = 4-20 = -16 < 0
D > 0 (Positive)

🎯 Two different real roots

\[x = \frac{-b \pm \sqrt{D}}{2a}\]

Since √D is real and positive, we get two distinct values

Example:

x² - 7x + 10 = 0

D = 49 - 40 = 9 > 0

Roots: x = 5, 2

D = 0 (Zero)

🎯 Two equal real roots

\[x = \frac{-b}{2a}\]

Both roots are same (coincident roots)

Example:

x² - 4x + 4 = 0

D = 16 - 16 = 0

Roots: x = 2, 2

D < 0 (Negative)

🎯 No real roots (complex)

\[x = \frac{-b \pm i\sqrt{|D|}}{2a}\]

Roots are complex conjugates

Example:

x² + 2x + 5 = 0

D = 4 - 20 = -16 < 0

Roots: -1 ± 2i

📝 Solved Example 4 (JEE Main 2023 Pattern)

Question: Find the value of k for which the equation kx² + 4x + 1 = 0 has real and equal roots.

Solution:

Given: kx² + 4x + 1 = 0

For real and equal roots: D = 0

Step 1: Identify coefficients

a = k, b = 4, c = 1

Step 2: Apply condition D = 0

b² - 4ac = 0

(4)² - 4(k)(1) = 0

16 - 4k = 0

4k = 16

k = 4

Verification:

When k = 4: 4x² + 4x + 1 = 0

D = 16 - 16 = 0 ✓

Answer: k = 4

3.3 Conditions for Different Types of Roots

Quick Reference Table - Root Conditions

Required Root Type Condition on D Additional Info
Real roots D ≥ 0 Both equal or unequal
Real and distinct D > 0 Two different real numbers
Real and equal D = 0 Perfect square trinomial
Rational roots D ≥ 0 & D is perfect square a, b, c must be rational
Irrational roots D > 0 & D not perfect square Occur in conjugate pairs
Imaginary/Complex D < 0 Complex conjugate pairs

⚠️ Common Mistakes with Discriminant

  • Sign errors in b² - 4ac: Be careful with negative values of b and c
  • Forgetting to check perfect square: For rational roots, D must be perfect square
  • D = 0 interpretation: This gives ONE root repeated twice, not one root only
  • Real vs Rational: All rational numbers are real, but not all real are rational
  • Coefficient identification: Always write in standard form first
4

Vieta's Formulas (Sum and Product of Roots)

Vieta's formulas establish relationships between roots and coefficients WITHOUT solving the equation. This is one of the MOST IMPORTANT concepts for JEE.

4.1 Vieta's Formulas - The Golden Rules

⭐ MOST IMPORTANT FORMULAS

For quadratic equation ax² + bx + c = 0 with roots α and β:

Sum of Roots:

\[\alpha + \beta = -\frac{b}{a}\]

Product of Roots:

\[\alpha \cdot \beta = \frac{c}{a}\]

Memory Trick:

✅ Sum = -b/a (coefficient of x with opposite sign ÷ coefficient of x²)

✅ Product = c/a (constant term ÷ coefficient of x²)

For monic equation (a = 1): x² + px + q = 0

Sum = -p, Product = q

📝 Solved Example 5 - Basic Vieta's

Question: If α and β are roots of 3x² - 7x + 2 = 0, find:
(a) α + β
(b) αβ
(c) α² + β²
(d) 1/α + 1/β

Solution:

Given: 3x² - 7x + 2 = 0

Coefficients: a = 3, b = -7, c = 2

(a) Sum of roots: α + β

α + β = -b/a = -(-7)/3 = 7/3

α + β = 7/3

(b) Product of roots: αβ

αβ = c/a = 2/3

αβ = 2/3

(c) α² + β²

Use identity: α² + β² = (α + β)² - 2αβ

= (7/3)² - 2(2/3)

= 49/9 - 4/3

= 49/9 - 12/9 = 37/9

α² + β² = 37/9

(d) 1/α + 1/β

1/α + 1/β = (β + α)/(αβ)

= (α + β)/(αβ)

= (7/3)/(2/3)

= 7/2

1/α + 1/β = 7/2

4.2 Important Symmetric Functions of Roots

Derived Formulas (Must Memorize for JEE)

Expression Formula In terms of a, b, c
α + β Given -b/a
αβ Given c/a
α² + β² (α+β)² - 2αβ (b² - 2ac)/a²
α³ + β³ (α+β)³ - 3αβ(α+β) (-b³ + 3abc)/a³
α - β ±√[(α+β)² - 4αβ] ±√D/a
1/α + 1/β (α+β)/(αβ) -b/c
1/α² + 1/β² (α²+β²)/(αβ)² (b² - 2ac)/c²
|α - β| √D/|a| √(b² - 4ac)/|a|

💡 JEE Shortcuts for Vieta's Formulas

Quick Pattern Recognition:

  • α² + β² uses (sum)² - 2(product)
  • α³ + β³ uses (sum)³ - 3(product)(sum)
  • 1/α + 1/β = sum/product
  • α - β needs discriminant

Common JEE Tricks:

  • If roots equal, use D = 0
  • If one root is reciprocal: c/a = 1
  • If roots opposite sign: c/a < 0
  • If roots same sign: c/a > 0
5

Formation of Quadratic Equations

⭐ Formation Formula

If α and β are roots, the quadratic equation is:

\[x^2 - (\alpha + \beta)x + \alpha\beta = 0\]

OR

\[x^2 - (Sum\ of\ roots)x + (Product\ of\ roots) = 0\]

📝 Solved Example 6

Question: Form a quadratic equation whose roots are 3 and -5.

Given: α = 3, β = -5

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

Product of roots = 3 × (-5) = -15

Equation: x² - (sum)x + (product) = 0

x² - (-2)x + (-15) = 0

x² + 2x - 15 = 0

6

Graph and Parabola

The graph of y = ax² + bx + c is a parabola. Understanding parabola properties is crucial for JEE Advanced.

Key Features of Parabola

Vertex (Turning Point):

\[V = \left(-\frac{b}{2a}, -\frac{D}{4a}\right)\]

Axis of Symmetry:

\[x = -\frac{b}{2a}\]

Maximum/Minimum Value:

\[y_{extrema} = -\frac{D}{4a}\]

Direction:

a > 0: Opens upward (∪)

a < 0: Opens downward (∩)

🎯 Range of Quadratic Function

If a > 0: Range = [-D/4a, ∞) [Minimum at vertex]

If a < 0: Range = (-∞, -D/4a] [Maximum at vertex]

7

Quadratic Inequalities

Solution of ax² + bx + c > 0 (or < 0)

Step 1: Find roots (if they exist)

Step 2: Draw sign diagram

Step 3: Check sign in intervals

Wavy Curve Method:

  • Mark roots on number line
  • Start from right (+ if a > 0, - if a < 0)
  • Alternate signs at each root
  • Choose intervals based on inequality
8

Applications and Advanced Concepts

8.1 Common Roots Problems

If equations ax² + bx + c = 0 and px² + qx + r = 0 have one common root α:

\[(ar - cp)^2 = (aq - bp)(br - cq)\]

8.2 Important Results

Condition Relationship
Roots are reciprocal a = c
Roots are equal in magnitude but opposite in sign b = 0
Both roots positive D ≥ 0, -b/a > 0, c/a > 0
Both roots negative D ≥ 0, -b/a < 0, c/a > 0
Roots opposite in sign c/a < 0

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

  • ✓ Discriminant & Nature of Roots: 35%
  • ✓ Vieta's Formulas Applications: 30%
  • ✓ Formation & Common Roots: 20%
  • ✓ Graph & Range Problems: 15%

JEE Advanced (Last 5 Years)

  • ✓ Quadratic Inequalities: 30%
  • ✓ Range & Max-Min Problems: 25%
  • ✓ Conditions on roots: 25%
  • ✓ Complex roots & applications: 20%

Top 15 Most Repeated Question Types

  1. Finding value of k for given nature of roots
  2. Sum and product of roots problems
  3. Formation of equation with transformed roots
  4. Common roots between two equations
  5. Range of quadratic expression
  6. Maximum/minimum value problems
  7. Quadratic inequalities solution
  8. Symmetric functions of roots
  9. Conditions for positive/negative roots
  10. Graph interpretation questions
  11. Reciprocal roots conditions
  12. Roots in given interval
  13. Sign of quadratic expression
  14. Integer/rational roots conditions
  15. Application word problems

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)
JEE Advanced: 9-12 marks (2-3 questions)
Difficulty Level: Easy to Hard (Mixed)
Time Required for Prep: 4-5 hours
Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Foundation)

  1. Solve: x² - 7x + 12 = 0 by factorization
  2. Find discriminant and nature of roots: 2x² + 5x + 3 = 0
  3. If α and β are roots of x² - 5x + 6 = 0, find α + β and αβ
  4. Form equation with roots 3 and -4
  5. Solve: x² - 16 = 0
  6. Find value of k if x² + kx + 9 = 0 has equal roots
  7. Solve using quadratic formula: 3x² - 5x + 2 = 0
  8. Find sum of squares of roots of x² - 6x + 5 = 0

Level 2: Intermediate (JEE Main Standard)

  1. Find range of k for which x² + kx + 4 = 0 has real roots
  2. If one root is twice the other in ax² + bx + c = 0, prove 2b² = 9ac
  3. Solve: (x - 1)(x - 2) = (x - 3)(x - 4)
  4. Find α² + β² if α, β are roots of 2x² - 3x + 5 = 0
  5. For what value of k do x² - 4x + k = 0 and x² - kx + 4 = 0 have one common root?
  6. Find maximum value of -2x² + 8x - 5
  7. Solve: |x² - 5x + 6| = x² - 5x + 6
  8. If α, β are roots of x² - px + q = 0, find equation whose roots are α² and β²

Level 3: Advanced (JEE Advanced Pattern)

  1. Find all values of a for which both roots of x² - 2ax + a² - 1 = 0 lie in (-2, 2)
  2. If α, β are roots of x² - 3x + a = 0 and γ, δ are roots of x² - 12x + b = 0, and α, β, γ, δ form an increasing GP, find a and b
  3. Solve for x: √(x² - 3x + 2) + √(x² - 3x - 4) = 4
  4. Find range of f(x) = (x² + x + 1)/(x² - x + 1)
  5. If x² - (m - 3)x + m = 0 has roots α, β such that α² + β² = 6, find m
  6. Solve: x²/(x - 1) + (x - 1)/x² = 5/2
  7. Find number of integral values of k for which x² - 2(k + 1)x + k² = 0 has both roots between -5 and 5
  8. If α, β are roots of ax² + bx + c = 0, find equation whose roots are (α + 1/β) and (β + 1/α)
Get Solutions PDF with Step-by-Step Working

Quadratic Equations - Complete Guide for JEE 2025-26

Why Quadratic Equations is Crucial for JEE?

Quadratic Equations is one of the highest weightage topics in JEE Mathematics, carrying 4-5% direct marks:

  • Complex Numbers - Foundation for imaginary roots
  • Functions - Quadratic functions are extensively tested
  • Sequences & Series - GP/AP problems use quadratics
  • Coordinate Geometry - Parabola and conic sections
  • Calculus - Maxima-minima applications

Most Important Formulas to Remember

1. Quadratic Formula

x = [-b ± √(b² - 4ac)]/(2a)

2. Discriminant

D = b² - 4ac

3. Vieta's Formulas

α + β = -b/a

αβ = c/a

4. Vertex

x = -b/(2a), y = -D/(4a)

📚 How to Study Quadratic Equations Effectively?

For JEE Main Students:

  1. Time Required: 3-4 days (3 hours/day)
  2. Master discriminant concept thoroughly
  3. Practice 100+ Vieta's formula problems
  4. Learn all factorization tricks
  5. Focus on nature of roots questions (very common)

For JEE Advanced Students:

  1. Time Required: 4-5 days (4 hours/day)
  2. Master quadratic inequalities
  3. Practice range and domain problems extensively
  4. Study conditions on roots in detail
  5. Solve complex application problems

Related Mathematics Notes

Complex Numbers

Imaginary numbers, Argand plane, Euler's formula

Sequences & Series

AP, GP, HP, AGP, Sum to n terms

Functions

Types, Domain, Range, Composite, Inverse