Quadratic Equations

Class 10 NCERT Mathematics - Chapter 4

Important Notes from Exercises

What is a Quadratic Equation?

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.

Standard Form

ax² + bx + c = 0, where a ≠ 0

a = coefficient of x²

b = coefficient of x

c = constant term

Key Concepts from Exercise 4.1

Checking if an equation is quadratic:

1. Expand and simplify the equation

2. Arrange in standard form: ax² + bx + c = 0

3. Check if highest power of x is 2 and a ≠ 0

4. If highest power > 2 → NOT quadratic

5. If a = 0 after simplification → NOT quadratic (becomes linear)

Key Concepts from Exercise 4.2

Factorization Method

Step 1: Write equation in standard form: ax² + bx + c = 0

Step 2: Split middle term (b) into two numbers whose:

Product = a × c

Sum = b

Step 3: Factor by grouping

Step 4: Apply zero product property

Tips for Factorization:

For equations with √2, √3 etc., treat them as coefficients

For fractional coefficients, multiply by LCM to clear denominators first

Perfect square trinomials: a² ± 2ab + b² = (a ± b)²

Difference of squares: a² - b² = (a+b)(a-b)

Key Concepts from Exercise 4.3

Discriminant (D) = b² - 4ac

D > 0 → Two distinct real roots

D = 0 → Two equal real roots (perfect square)

D < 0 → No real roots (imaginary roots)

Quadratic Formula

x = [-b ± √(b² - 4ac)] / 2a

Condition for Equal Roots:

For equal roots: D = 0, i.e., b² - 4ac = 0

This gives: b² = 4ac

Use this to find unknown coefficients like k

Word Problem Strategy:

1. Identify the unknown → Let it be x

2. Express other quantities in terms of x

3. Form the equation using given condition

4. Simplify to standard quadratic form

5. Solve and verify the answer makes sense

Common Word Problem Patterns:

Area problems: Length × Breadth = Area

Age problems: Product/Sum of ages = given value

Speed problems: Speed = Distance/Time

Number problems: Consecutive integers, sum/product

Right triangle: Pythagoras theorem a² + b² = c²

Exercise 4.1 - Introduction to Quadratic Equations

Question 1: Check whether the following are quadratic equations

(i) (x + 1)² = 2(x - 3) Easy

Solution:

Given: (x + 1)² = 2(x - 3)
Step 1: Expand LHS → x² + 2x + 1 = 2x - 6
Step 2: Bring all terms to one side → x² + 2x + 1 - 2x + 6 = 0
Step 3: Simplify → x² + 7 = 0
This is of the form ax² + bx + c = 0 where a=1, b=0, c=7
It IS a Quadratic Equation
(ii) x² - 2x = (-2)(3 - x) Easy

Solution:

Given: x² - 2x = (-2)(3 - x)
Step 1: Expand RHS → x² - 2x = -6 + 2x
Step 2: Bring all terms to LHS → x² - 2x + 6 - 2x = 0
Step 3: Simplify → x² - 4x + 6 = 0
This is of the form ax² + bx + c = 0 where a=1, b=-4, c=6
It IS a Quadratic Equation
(iii) (x - 2)(x + 1) = (x - 1)(x + 3) Medium

Solution:

Given: (x - 2)(x + 1) = (x - 1)(x + 3)
Step 1: Expand LHS → x² + x - 2x - 2 = x² - x - 2
Step 2: Expand RHS → x² + 3x - x - 3 = x² + 2x - 3
Step 3: Equate → x² - x - 2 = x² + 2x - 3
Step 4: Cancel x² from both sides → -x - 2 = 2x - 3
Step 5: Simplify → -3x + 1 = 0 or 3x - 1 = 0
This is a linear equation (degree 1), NOT quadratic
It is NOT a Quadratic Equation
(iv) (x - 3)(2x + 1) = x(x + 5) Easy

Solution:

Given: (x - 3)(2x + 1) = x(x + 5)
Step 1: Expand LHS → 2x² + x - 6x - 3 = 2x² - 5x - 3
Step 2: Expand RHS → x² + 5x
Step 3: Bring all terms to LHS → 2x² - 5x - 3 - x² - 5x = 0
Step 4: Simplify → x² - 10x - 3 = 0
This is of the form ax² + bx + c = 0 where a=1, b=-10, c=-3
It IS a Quadratic Equation
(v) (2x - 1)(x - 3) = (x + 5)(x - 1) Easy

Solution:

Given: (2x - 1)(x - 3) = (x + 5)(x - 1)
Step 1: Expand LHS → 2x² - 6x - x + 3 = 2x² - 7x + 3
Step 2: Expand RHS → x² - x + 5x - 5 = x² + 4x - 5
Step 3: Bring all terms to LHS → 2x² - 7x + 3 - x² - 4x + 5 = 0
Step 4: Simplify → x² - 11x + 8 = 0
This is of the form ax² + bx + c = 0 where a=1, b=-11, c=8
It IS a Quadratic Equation
(vi) x² + 3x + 1 = (x - 2)² Easy

Solution:

Given: x² + 3x + 1 = (x - 2)²
Step 1: Expand RHS → x² - 4x + 4
Step 2: Bring all terms to LHS → x² + 3x + 1 - x² + 4x - 4 = 0
Step 3: Simplify → 7x - 3 = 0
The x² terms cancel out! This is a linear equation (degree 1)
It is NOT a Quadratic Equation
(vii) (x + 2)³ = 2x(x² - 1) Hard

Solution:

Given: (x + 2)³ = 2x(x² - 1)
Step 1: Expand LHS using (a+b)³ = a³ + 3a²b + 3ab² + b³
→ x³ + 6x² + 12x + 8
Step 2: Expand RHS → 2x³ - 2x
Step 3: Bring all terms to LHS → x³ + 6x² + 12x + 8 - 2x³ + 2x = 0
Step 4: Simplify → -x³ + 6x² + 14x + 8 = 0
Highest power of x is 3, so this is a CUBIC equation
It is NOT a Quadratic Equation
(viii) x³ - 4x² - x + 1 = (x - 2)³ Hard

Solution:

Given: x³ - 4x² - x + 1 = (x - 2)³
Step 1: Expand RHS using (a-b)³ = a³ - 3a²b + 3ab² - b³
→ x³ - 6x² + 12x - 8
Step 2: Bring all terms to LHS → x³ - 4x² - x + 1 - x³ + 6x² - 12x + 8 = 0
Step 3: Simplify → 2x² - 13x + 9 = 0
The x³ terms cancel out! Highest power is 2
This is of the form ax² + bx + c = 0 where a=2, b=-13, c=9
It IS a Quadratic Equation

Question 2: Represent situations as quadratic equations

(i) Area of rectangular plot is 528 m². Length is one more than twice its breadth. Find length and breadth. Medium

Solution:

Let breadth = x metres
Then length = (2x + 1) metres (one more than twice breadth)
Area = Length × Breadth = 528
→ (2x + 1) × x = 528
→ 2x² + x = 528
→ 2x² + x - 528 = 0
Solving: x = 16 (breadth), Length = 33 m
(ii) Product of two consecutive positive integers is 306. Find the integers. Easy

Solution:

Let first integer = x
Then second integer = (x + 1) (consecutive)
Product = 306
→ x(x + 1) = 306
→ x² + x = 306
→ x² + x - 306 = 0
Solving: x = 17, so integers are 17 and 18
(iii) Rohan's mother is 26 years older. Product of their ages 3 years from now is 360. Find Rohan's present age. Medium

Solution:

Let Rohan's present age = x years
Mother's present age = (x + 26) years
3 years from now:
Rohan's age = (x + 3), Mother's age = (x + 29)
Product = 360
→ (x + 3)(x + 29) = 360
→ x² + 29x + 3x + 87 = 360
→ x² + 32x + 87 - 360 = 0
→ x² + 32x - 273 = 0
Solving: x = 7 (Rohan's age = 7 years)
(iv) Train travels 480 km. If speed were 8 km/h less, it would take 3 hours more. Find speed. Hard

Solution:

Let speed of train = x km/h
Time taken = 480/x hours
Reduced speed = (x - 8) km/h
New time = 480/(x-8) hours
Difference in time = 3 hours
→ 480/(x-8) - 480/x = 3
→ 480[x - (x-8)] / [x(x-8)] = 3
→ 480 × 8 / (x² - 8x) = 3
→ 3840 = 3(x² - 8x)
→ 1280 = x² - 8x
→ x² - 8x - 1280 = 0
Solving: x = 40 km/h (speed of train)
Exercise 4.2 - Solving by Factorization

Question 1: Find roots by factorization

(i) x² - 3x - 10 = 0 Easy

Solution:

Given: x² - 3x - 10 = 0
Step 1: Find two numbers whose product = -10 and sum = -3
Numbers: -5 and +2 (since -5 × 2 = -10 and -5 + 2 = -3)
Step 2: Split middle term → x² - 5x + 2x - 10 = 0
Step 3: Group → x(x - 5) + 2(x - 5) = 0
Step 4: Factor → (x - 5)(x + 2) = 0
Step 5: x - 5 = 0 or x + 2 = 0
Roots: x = 5 and x = -2
(ii) 2x² + x - 6 = 0 Medium

Solution:

Given: 2x² + x - 6 = 0
Step 1: Find two numbers whose product = 2×(-6) = -12 and sum = 1
Numbers: +4 and -3 (since 4 × (-3) = -12 and 4 + (-3) = 1)
Step 2: Split middle term → 2x² + 4x - 3x - 6 = 0
Step 3: Group → 2x(x + 2) - 3(x + 2) = 0
Step 4: Factor → (x + 2)(2x - 3) = 0
Step 5: x + 2 = 0 or 2x - 3 = 0
Roots: x = -2 and x = 3/2
(iii) √2 x² + 7x + 5√2 = 0 Hard

Solution:

Given: √2 x² + 7x + 5√2 = 0
Step 1: Find two numbers whose product = √2 × 5√2 = 10 and sum = 7
Numbers: 5 and 2 (since 5 × 2 = 10 and 5 + 2 = 7)
Step 2: Split middle term → √2 x² + 5x + 2x + 5√2 = 0
Step 3: Group → x(√2x + 5) + √2(√2x + 5) ... Check: √2 × √2 = 2, not 5. Let me try differently
Correct approach: √2 x² + 2x + 5x + 5√2 = 0
→ √2x(x + √2) + 5(x + √2) ... Check: 5 × √2 = 5√2 ✓
→ (√2x + 5)(x + √2) = 0
→ x = -5/√2 or x = -√2
Roots: x = -√2 and x = -5/√2
Rationalized: x = -√2 and x = -5√2/2
(iv) 2x² - x + 1/8 = 0 Medium

Solution:

Given: 2x² - x + 1/8 = 0
Step 1: Multiply by 8 to clear fraction → 16x² - 8x + 1 = 0
Step 2: Find two numbers whose product = 16×1 = 16 and sum = -8
Numbers: -4 and -4 (since (-4)×(-4) = 16 and -4+(-4) = -8)
Step 3: Split middle term → 16x² - 4x - 4x + 1 = 0
Step 4: Group → 4x(4x - 1) - 1(4x - 1) = 0
Step 5: Factor → (4x - 1)(4x - 1) = 0
Step 6: (4x - 1)² = 0
→ 4x - 1 = 0 → x = 1/4
Roots: x = 1/4, 1/4 (Equal roots)
(v) 100x² - 20x + 1 = 0 Medium

Solution:

Given: 100x² - 20x + 1 = 0
Step 1: Find two numbers whose product = 100×1 = 100 and sum = -20
Numbers: -10 and -10 (since (-10)×(-10) = 100 and -10+(-10) = -20)
Step 2: Split middle term → 100x² - 10x - 10x + 1 = 0
Step 3: Group → 10x(10x - 1) - 1(10x - 1) = 0
Step 4: Factor → (10x - 1)(10x - 1) = 0
Step 5: (10x - 1)² = 0
→ 10x - 1 = 0 → x = 1/10
Roots: x = 1/10, 1/10 (Equal roots)
This is a perfect square: (10x - 1)² = 0

Question 3: Find two numbers whose sum is 27 and product is 182

Find two numbers with sum 27 and product 182 Easy

Solution:

Let the numbers be x and (27 - x)
Product = x(27 - x) = 182
→ 27x - x² = 182
→ x² - 27x + 182 = 0
Find two numbers whose product = 182 and sum = 27
Numbers: 13 and 14 (since 13 × 14 = 182 and 13 + 14 = 27)
→ x² - 13x - 14x + 182 = 0
→ x(x - 13) - 14(x - 13) = 0
→ (x - 13)(x - 14) = 0
Numbers are 13 and 14

Question 4: Find two consecutive positive integers, sum of squares is 365

Two consecutive positive integers, sum of squares = 365 Medium

Solution:

Let the integers be x and (x + 1)
Sum of squares = x² + (x+1)² = 365
→ x² + x² + 2x + 1 = 365
→ 2x² + 2x + 1 - 365 = 0
→ 2x² + 2x - 364 = 0
→ x² + x - 182 = 0 (dividing by 2)
Find two numbers whose product = -182 and sum = 1
Numbers: 14 and -13 (since 14 × (-13) = -182 and 14 + (-13) = 1)
→ x² + 14x - 13x - 182 = 0
→ x(x + 14) - 13(x + 14) = 0
→ (x + 14)(x - 13) = 0
x = -14 or x = 13 (taking positive)
Integers are 13 and 14
Check: 13² + 14² = 169 + 196 = 365 ✓

Question 5: Right triangle - altitude 7 cm less than base, hypotenuse 13 cm

Find the other two sides of the right triangle Medium

Solution:

Let base = x cm
Then altitude = (x - 7) cm
Hypotenuse = 13 cm
By Pythagoras: base² + altitude² = hypotenuse²
→ x² + (x - 7)² = 13²
→ x² + x² - 14x + 49 = 169
→ 2x² - 14x + 49 - 169 = 0
→ 2x² - 14x - 120 = 0
→ x² - 7x - 60 = 0 (dividing by 2)
Find two numbers whose product = -60 and sum = -7
Numbers: -12 and 5 (since -12 × 5 = -60 and -12 + 5 = -7)
→ x² - 12x + 5x - 60 = 0
→ x(x - 12) + 5(x - 12) = 0
→ (x - 12)(x + 5) = 0
x = 12 or x = -5 (taking positive)
Base = 12 cm, Altitude = 12 - 7 = 5 cm
Other two sides: 5 cm and 12 cm
Check: 5² + 12² = 25 + 144 = 169 = 13² ✓

Question 6: Cottage industry - cost and production problem

Find number of articles and cost per article Hard

Solution:

Let number of articles = x
Cost per article = (2x + 3) rupees (3 more than twice the number)
Total cost = Number × Cost per article = 90
→ x(2x + 3) = 90
→ 2x² + 3x = 90
→ 2x² + 3x - 90 = 0
Find two numbers whose product = 2×(-90) = -180 and sum = 3
Numbers: 15 and -12 (since 15 × (-12) = -180 and 15 + (-12) = 3)
→ 2x² + 15x - 12x - 90 = 0
→ x(2x + 15) - 6(2x + 15) = 0
→ (2x + 15)(x - 6) = 0
x = -15/2 or x = 6 (taking positive integer)
Number of articles = 6
Cost per article = 2(6) + 3 = Rs 15
6 articles at Rs 15 each
Check: 6 × 15 = 90 ✓
Exercise 4.3 - Nature of Roots & Quadratic Formula

Question 1: Find nature of roots. If real roots exist, find them.

(i) 2x² - 3x + 5 = 0 Easy

Solution:

Given: 2x² - 3x + 5 = 0
Compare with ax² + bx + c = 0
a = 2, b = -3, c = 5
Discriminant D = b² - 4ac
D = (-3)² - 4(2)(5) = 9 - 40 = -31
Since D = -31 < 0
No real roots exist (roots are imaginary)
Roots: x = [3 ± √(-31)] / 4 = [3 ± i√31] / 4
(ii) 3x² - 4√3 x + 4 = 0 Medium

Solution:

Given: 3x² - 4√3 x + 4 = 0
a = 3, b = -4√3, c = 4
D = b² - 4ac = (-4√3)² - 4(3)(4)
D = 16 × 3 - 48 = 48 - 48 = 0
Since D = 0, roots are real and equal
x = [-b ± √D] / 2a = [4√3 ± 0] / 6 = 4√3 / 6 = 2√3 / 3 = 2/√3
Two equal roots: x = 2/√3 (or 2√3/3)
(iii) 2x² - 6x + 3 = 0 Medium

Solution:

Given: 2x² - 6x + 3 = 0
a = 2, b = -6, c = 3
D = b² - 4ac = (-6)² - 4(2)(3) = 36 - 24 = 12
Since D = 12 > 0, two distinct real roots exist
x = [6 ± √12] / 4 = [6 ± 2√3] / 4 = [3 ± √3] / 2
Roots: x = (3 + √3)/2 and x = (3 - √3)/2
Approximate: x ≈ 2.366 and x ≈ 0.634

Question 2: Find values of k for equal roots

(i) 2x² + kx + 3 = 0 Medium

Solution:

Given: 2x² + kx + 3 = 0
For equal roots: D = b² - 4ac = 0
a = 2, b = k, c = 3
D = k² - 4(2)(3) = 0
→ k² - 24 = 0
→ k² = 24
→ k = ±√24 = ±2√6
k = ±2√6 (or ±√24)
(ii) kx(x - 2) + 6 = 0 Hard

Solution:

Given: kx(x - 2) + 6 = 0
→ kx² - 2kx + 6 = 0
a = k, b = -2k, c = 6
For equal roots: D = b² - 4ac = 0
D = (-2k)² - 4(k)(6) = 0
→ 4k² - 24k = 0
→ 4k(k - 6) = 0
→ k = 0 or k = 6
But k = 0 makes it linear (not quadratic), so k ≠ 0
k = 6

Question 3: Rectangular mango grove - length twice breadth, area 800 m²

Is it possible? If so, find length and breadth. Medium

Solution:

Let breadth = x m, then length = 2x m
Area = Length × Breadth = 800
→ 2x × x = 800
→ 2x² = 800
→ x² = 400
→ x = ±20 (taking positive)
Breadth = 20 m, Length = 40 m
Yes, it is possible! Length = 40 m, Breadth = 20 m
Check: 40 × 20 = 800 ✓

Question 4: Sum of ages of two friends is 20. Four years ago, product was 48.

Is this possible? If so, find their present ages. Hard

Solution:

Let present age of one friend = x years
Then present age of other = (20 - x) years
Four years ago:
Their ages were (x - 4) and (16 - x)
Product = 48
→ (x - 4)(16 - x) = 48
→ 16x - x² - 64 + 4x = 48
→ -x² + 20x - 64 = 48
→ -x² + 20x - 112 = 0
→ x² - 20x + 112 = 0
D = (-20)² - 4(1)(112) = 400 - 448 = -48
Since D = -48 < 0, no real solution exists
This situation is NOT possible

Question 5: Rectangular park of perimeter 80 m and area 400 m²

Is it possible? If so, find length and breadth. Medium

Solution:

Let length = x m, breadth = y m
Perimeter = 2(x + y) = 80 → x + y = 40 → y = 40 - x
Area = x × y = 400
→ x(40 - x) = 400
→ 40x - x² = 400
→ x² - 40x + 400 = 0
D = (-40)² - 4(1)(400) = 1600 - 1600 = 0
Since D = 0, equal roots exist
x = [40 ± 0] / 2 = 20
Length = 20 m, Breadth = 20 m
Yes, it is possible! It's a square with sides 20 m
Check: Perimeter = 2(20+20) = 80 ✓, Area = 20×20 = 400 ✓

Scientific Calculator

0

Quadratic Equation Solver

Enter coefficients for ax² + bx + c = 0

Quadratic Equation Visualizer

Visualize the parabola y = ax² + bx + c