Quadratic Equations for Class 10: Important Questions and Step-by-Step Solutions

A quadratic equation is generally expressed in the standard form ax²+bx+c=0, where a ≠ 0. This chapter introduces students to various methods of solving quadratic equations, including factorisation, completing the square, and the quadratic formula. It also explains the nature of roots through the discriminant. In this guide, you will find carefully selected important questions covering all major concepts from the Class 10 Quadratic Equations chapter, designed according to the latest syllabus and exam pattern to support effective preparation and revision.

Our subject experts have provided detailed solutions for these problems based on the latest CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• Key Concepts of the Quadratic Equation for Class 10
• Section A: 1-Mark Questions (MCQ & VSA)
• Section B: 2-Mark Questions (Short Answer)
• Section C: 3-Mark Questions
• Section D: Word Problems
• Section E: 4-Mark & Case Study Questions
• Previous Year Questions

Key Concepts of the Quadratic Equation for Class 10

A polynomial equation of degree 2 in one variable, where the coefficient of x² is non-zero. ax² + bx + c = 0 ,a ≠ 0

• Roots / Solutions: 

Roots or solutions are values of x that satisfy the equation. A quadratic equation has exactly two roots (which may be equal, real, or imaginary).

p(α) = 0 ⇒  α is a root

• Discriminant (D):

The expression b² − 4ac determines the nature of the roots without actually solving the equation.

D = b² − 4ac

• Quadratic Formula:

For a quadratic equation ax² + bx + c = 0 ,a ≠ 0

x = [−b ± √(b²−4ac)] / 2a

• Completing the Square: 

Converts any quadratic into the form (x + k)² = n, making it easy to extract the roots.

x² + bx = −c

x² + bx + (b/2)² = (b/2)² − c

• Sum & Product of Roots:

For roots α and β of ax² + bx + c = 0, a direct relationship exists with the coefficients.

α + β = −b/a

αβ = c/a

• Nature of roots

The discriminant D = b² − 4ac determines the nature of the quadratic equation ax² + bx + c = 0.

If D > 0: Two distinct real roots

If D = 0: Two equal real roots (repeated roots)

If D < 0: No real roots (imaginary roots only)

Section A: 1-Mark Questions (MCQ & VSA)

Q1: The discriminant of the quadratic equation bx² + ax + c = 0 (b ≠ 0) is:

(a) b² − 4ac (b) b² − 4bc

(c) a² − 4bc (d) a² − 4bc 

Solution: Compare bx² + ax + c with Ax² + Bx + C. Here A = b, B = a, C = c. So D = B² − 4AC = a² − 4bc.

Answer: (c) a² − 4bc

Q2: Which of the following quadratic equations has real and equal roots?

(a) (x + 1)²  (b) x² + x = 0

(c) x² − 4 = 0 (d) x² + x + 1 = 0

Solution: Expand (a): x² + 2x + 1 = 2x + 1 

⇒ x² = 0 

⇒ x = 0, 0. Equal roots.

Also, D = 0.

Answer: (a) (x + 1)²

Q3: For what value of k does kx² − 6x − 2 = 0 have equal roots?

(a) k = −9/2 (b) k = 9/2 

(c) k = 2/9 (d) k = −2/9

Solution: For equal roots, D = 0.

D = b² − 4ac = 36 − 4(k)(−2) = 36 + 8k = 0 

⇒ k = −9/2.

Answer: (a) k = −9/2

Q4: Check whether x = −2 is a solution of the equation x² + 3x + 2 = 0.

Solution: Substitute x = −2: (−2)² + 3(−2) + 2 = 4 − 6 + 2 = 0 

x = −2 is a solution of the equation x² + 3x + 2 = 0.

Q5: Write the discriminant of the equation 3x² − 2x + 1/3 = 0.

Solution: a = 3, b = −2, c = 1/3. 

D = b² − 4ac = 4 − 4(3)(1/3) = 4 − 4 = 0

Discriminant = 0 (equal real roots)

Q6: Is x² − 1 = (x − 1)(x + 1) a quadratic equation?

Solution: Expanding: x² − 1 = x² − 1. Both sides cancel to 0 = 0, which has no variable. This is an identity, not an equation. So it is not a quadratic equation.

Section B: 2-Mark Questions (Short Answer)

Q7: Solve the quadratic equation 2x² + x − 6 = 0 by factorisation.

Solution: Find two numbers whose product = 2 × (−6) = −12 and sum = 1. The numbers are 4 and −3.

2x² + 4x − 3x − 6 = 0 

⇒  2x(x + 2) − 3(x + 2) = 0 

⇒  (2x − 3)(x + 2) = 0

Therefore x = 3/2 or x = −2

Roots: x = 3/2 and x = −2

Q8: Find the nature of roots of the quadratic equation 2x² − 4x + 3 = 0.

Solution: a = 2, b = −4, c = 3. 

D = b² − 4ac = 16 − 4(2)(3) = 16 − 24 = −8

Since D < 0, the equation has no real roots.

No real roots exist for the quadratic equation 2x² − 4x + 3 = 0(D = −8 < 0).

Q9: Find the value of k so that the equation x² − 4x + k = 0 has equal roots.

Solution: For equal roots, D = 0. Here a = 1, b = −4, c = k.

D = 16 − 4(1)(k) = 0

⇒  16 = 4k 

⇒  k = 4

Q10: Solve: √2 x² + 7x + 5√2 = 0

Solution: Product = √2 × 5√2 = 10, Sum = 7. Two numbers: 5 and 2.

√2 x² + 5x + 2x + 5√2 = 0 

⇒  x(√2 x + 5) + √2(√2 x + 5) = 0 

⇒  (x + √2)(√2 x + 5) = 0

x = −√2 or x = −5/√2 = −5√2/2

Roots: x = −√2 and x = −5√2/2

Q11: If one root of x² − 3x + k = 0 is 2, find the value of k and the other root.

Solution: Since 2 is a root: 4 − 6 + k = 0 

⇒ k = 2

Sum of roots = 3. So other root = 3 − 2 = 1

k = 2, Other root = 1

Section C: 3-Mark Questions

Q12: Solve x² − 5x + 6 = 0 using the method of completing the square.

Solution: x² − 5x + 6 = 0 

⇒  x² − 5x = −6

Add (5/2)² = 25/4 to both sides: x² − 5x + 25/4 = −6 + 25/4 = 1/4

(x − 5/2)² = 1/4 

⇒  x − 5/2 = ±1/2

x = 5/2 + 1/2 = 3  or  x = 5/2 − 1/2 = 2

Roots of x² − 5x + 6 = 0 are x = 3 and x = 2

Q13: Find the smallest value of p for which x² − 2(p+1)x + p² = 0 has real roots. Then find those roots.

Solution: For real roots, D ≥ 0

[−2(p+1)]² − 4(1)(p²) ≥ 0

4(p² + 2p + 1) − 4p² ≥ 0 

⇒  4p² + 8p + 4 − 4p² ≥ 0 

⇒  8p + 4 ≥ 0 

⇒  p ≥ −1/2

Smallest value of p = −1/2. 

Substituting back: x² − 2(−1/2 + 1)x + 1/4 = 0 

⇒  x² − x + 1/4 = 0 

⇒  (x − 1/2)² = 0

Both roots are x = 1/2 (equal roots).

Q14: Solve by the quadratic formula: 3x² − 5x + 2 = 0.

Solution: a = 3, b = −5, c = 2. 

D = 25 − 4(3)(2) = 25 − 24 = 1≥ 0

x = [−b ± √(b²−4ac)] / 2a =  [5 ± √1] / 6 = [5 ± 1] / 6

x = 6/6 = 1  or  x = 4/6 = 2/3

Roots of the quadratic equation 3x² − 5x + 2 = 0 are x = 1 and x = 2/3

Q15: Find the value(s) of k for which the equation (k+4)x² + (k+1)x + 1 = 0 has equal roots.

Solution: For equal roots, D = 0

(k+1)² − 4(k+4)(1) = 0

k² + 2k + 1 − 4k − 16 = 0 → k² − 2k − 15 = 0

(k − 5)(k + 3) = 0 → k = 5 or k = −3

Check k = −4: If k = −4, the equation becomes linear (coefficient of x² = 0). So k ≠ −4. Both k = 5 and k = −3 are valid.

k = 5 or k = −3

Section D: Word Problems

Q16: The length of a rectangle exceeds its breadth by 5 cm. If the area is 84 cm², find the length and breadth. (3 Marks)

Solution: Let breadth = x cm. Then length = (x + 5) cm.

Area = x(x + 5) = 84 

⇒  x² + 5x − 84 = 0

Factorise: (x + 12)(x − 7) = 0 

⇒  x = −12 (rejected, negative) or x = 7

Breadth = 7 cm, Length = 12 cm

Q17: A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and the speed of the boat in still water. (4 Marks)

Solution: Let speed of boat in still water = x km/h, speed of stream = y km/h.

Speed upstream = (x − y), Speed downstream = (x + y)

Equation 1: 30/(x−y) + 44/(x+y) = 10

Equation 2: 40/(x−y) + 55/(x+y) = 13

Let u = 1/(x−y) and v = 1/(x+y). 

Then: 30u + 44v = 10 and 40u + 55v = 13. 

Solving: u = 1/5 and v = 1/11 

⇒ x − y = 5 and x + y = 11 

⇒ x = 8, y = 3.

Speed of boat = 8 km/h, Speed of stream = 3 km/h

Q18: Seven years ago, Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two-fifth of Varun's age. Find their present ages. 

Solution: Let Swati's age 7 years ago = x years. Then Varun's age 7 years ago = 5x².

Present ages: Swati = (x + 7), Varun = (5x² + 7)

3 years hence: Swati = x + 10, Varun = 5x² + 10. 

Condition: (x + 10) = (2/5)(5x² + 10)

5(x + 10) = 2(5x² + 10)

⇒  5x + 50 = 10x² + 20 

⇒ 10x² − 5x − 30 = 0 

⇒ 2x² − x − 6 = 0

(2x + 3)(x − 2) = 0 

⇒ x = −3/2 (rejected) or x = 2

Swati's present age = 9 years, Varun's present age = 27 years

Q19: The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58/21, find the fraction. [CBSE 2024] 

Solution: Let numerator = x. Denominator = 2x + 1. 

Fraction = x/(2x + 1).

x/(2x+1) + (2x+1)/x = 58/21

Let t = x/(2x+1). 

Then t + 1/t = 58/21 

⇒ 21t² − 58t + 21 = 0 

⇒ 21t² − 49t − 9t + 21 = 0

(3t − 7)(7t − 3) = 0 

⇒ t = 7/3 or t = 3/7

t = 3/7 

⇒ x/(2x+1) = 3/7 

⇒ 7x = 6x + 3 

⇒ x = 3. Fraction = 3/7.

The fraction is 3/7

Section E: 4-Mark & Case Study Questions

Q20: Case Study: A garden designer plans a rectangular lawn surrounded by a uniform walkway of width x metres on all sides. The lawn is 12 m × 10 m. Total area of lawn + walkway = 360 m². Answer the sub-questions below. [CBSE 2025]

Solution:

(i) Formulate the quadratic equation:

With walkway width x, total length = 12 + 2x, total breadth = 10 + 2x.

(12 + 2x)(10 + 2x) = 360 

⇒  120 + 44x + 4x² = 360 

⇒  4x² + 44x − 240 = 0

Simplified: x² + 11x − 60 = 0

(ii) Solve for x:

x² + 15x − 4x − 60 = 0 

⇒  (x + 15)(x − 4) = 0 

⇒  x = −15 (rejected as width cannot be a negative quantity) or x = 4

(iii) Perimeter of the lawn:

Lawn is 12 m × 10 m. 

Perimeter = 2(12 + 10) = 44 m

Q21: There is a circular park of diameter 65 m. AB is a diameter. An entry gate P is on the boundary such that PA = PB + 35 m. Find PA and PB. [CBSE 2025]

Solution: Let PB = x m, then PA = (x + 35) m. 

Since AB is a diameter, ∠APB = 90° (angle in a semicircle).

By Pythagoras: AB² = PA² + PB² 

⇒  65² = (x + 35)² + x²

⇒ 4225 = x² + 70x + 1225 + x² 

⇒  2x² + 70x − 3000 = 0 

⇒  x² + 35x − 1500 = 0

(x + 60)(x − 25) = 0 

⇒  x = −60 (rejected as length cannot be negative) or x = 25

PB = 25 m, PA = 60 m

Q22: A student scored 32 marks in Maths and Science combined. Had he scored 4 more in Maths and 2 less in Science, the product of marks would be 253. Find his original marks. [CBSE 2025]

Solution: Let Maths marks = x. Then Science marks = (32 − x).

Condition: (x + 4)(32 − x − 2) = 253 

⇒  (x + 4)(30 − x) = 253

⇒ 30x − x² + 120 − 4x = 253 

⇒  −x² + 26x + 120 = 253 

⇒  x² − 26x + 133 = 0

(x − 19)(x − 7) = 0 

⇒  x = 19 or x = 7

Maths = 19, Science = 13  OR  Maths = 7, Science = 25

Previous Year Questions 

PYQ 2025 (Case Study)Q: A rectangular lawn (12 m × 10 m) is surrounded by a walkway of uniform width x. If total area = 360 m², find x.

Solution: Form equation: (12+2x)(10+2x) = 360

⇒ x² + 11x − 60 = 0 

⇒ (x+15)(x−4) = 0 

⇒  x = 4 (reject negative)

Answer: Width of walkway = 4 m

PYQ 2025 Q: Find the value(s) of k for which 4x² + kx + 1 = 0 has real and equal roots.

Solution: For equal roots, D = 0

 k² − 4(4)(1) = 0 

⇒  k² = 16 

⇒  k = ±4

Answer: k = +4 or k = −4

PYQ 2024 Q: In a flight of 2800 km, speed reduced by 100 km/h due to bad weather, increasing time by 30 min. Find original duration.

Solution: Let original speed = x. 

Time equations: 2800/(x−100) − 2800/x = 1/2. 

Solving: x² − 100x − 560000 = 0 

⇒  (x−800)(x+700) = 0 

⇒  x = 800 km/h. 

Duration = 2800/800 = 3.5 hours.

Answer: Original duration = 3 hours 30 minutes

PYQ 2024 Q: Denominator of a fraction is one more than twice the numerator. Sum of fraction and its reciprocal = 58/21. Find the fraction.

Solution: Let numerator = x, denominator = 2x+1. 

Solving x/(2x+1) + (2x+1)/x = 58/21 gives fraction = 3/7.

Answer: 3/7

PYQ 2023 Q: Two water taps together can fill a tank in 9 and 3/8 hours. The larger tap takes 10 hours less than the smaller one alone. Find the time each tap takes alone.

Solution: Let smaller tap take x hours. 

Larger takes (x−10). 

Combined: 1/x + 1/(x−10) = 8/75. 

Simplifying: 8x² − 230x + 750 = 0 

⇒  4x² − 115x + 375 = 0 

⇒  (x−25)(4x−15) = 0 

⇒  x = 25 (since x = 15/4 makes x−10 negative).

Smaller tap: 25 hours, Larger tap: 15 hours

PYQ 2022 Q: Solve by completing the square: 4x² + 4√3 x + 3 = 0. 

Solution: (2x)² + 2·(2x)·√3 + (√3)² = 0 

⇒  (2x + √3)² = 0 

⇒  2x = −√3 

⇒  x = −√3/2 (equal roots)

Answer: x = −√3/2 (both roots equal)

2021

PYQ 2021Q: If the roots of (a−b)x² + (b−c)x + (c−a) = 0 are equal, prove that 2a = b + c.

Solution: For equal roots, D = 0

(b−c)² − 4(a−b)(c−a) = 0. 

Expand: b² − 2bc + c² − 4(ac − a² − bc + ab) = 0. 

Simplify: (b+c−2a)² = 0 

⇒  b+c = 2a → 2a = b+c

Proved: 2a = b + c

PYQ 2020 Q: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed.

Solution: Let speed = x.

 360/x − 360/(x+5) = 1 

⇒ 360(x+5) − 360x = x(x+5) 

⇒ 1800 = x² + 5x 

⇒ x² + 5x − 1800 = 0 

⇒ (x+45)(x−40) = 0

 ⇒ x = 40.

Answer: Speed = 40 km/h