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Chapter 4 · Written-answer practice

Quadratic Equations 2-Mark Questions Class 10: CBSE PYQs with Answers

Practise 38 selected Quadratic Equations questions worth 2 marks each, from CBSE Basic and Standard board papers and official sample papers. Work on factorisation, the quadratic formula, discriminants, unknown coefficients and short word problems. Try each question before opening its solution.

38 distinct numbered questions · 64 board-paper appearances + 1 official-sample appearances
56 worked solutions, including both alternatives for all 18 OR pairs
Solutions prepared by Team TutorVertex
Updated

Key results

UseResult or method
Standard formax² + bx + c = 0, a ≠ 0. Here a, b and c are the coefficients after simplification.
DiscriminantD = b² − 4ac.
Nature of real rootsD > 0: two distinct real roots; D = 0: equal real roots; D < 0: no real roots.
Quadratic formulax = −b ± √D2a, for D ≥ 0.
Root sum and productα + β = −ba; αβ = ca.
Zero-product ruleIf uv = 0, then u = 0 or v = 0.

For a parameter question, check that the coefficient of x² stays non-zero. In a word problem, test each root against the stated conditions.

Which method should I use?
QuestionA useful starting point
Roots by factorisationBring all terms to one side, split the middle term or use a suitable identity, then set each factor equal to zero.
Quadratic formula is requestedIdentify a, b and c; calculate D; substitute in the formula and simplify both roots.
Only the nature of rootsCalculate D and state the conclusion from its sign.
Real and equal rootsSet D = 0, solve for the parameter and check a ≠ 0.
A root is givenSubstitute it into the original equation; use the root sum or product if another root is needed.
A short word problemDefine the unknown with units, form the quadratic equation and reject roots that fail the context.

Follow the method named in the question. These suggestions do not make a method compulsory when none is specified.

Showing all 38 questions

Questions on this page — Show list

Solving by factorisation

  1. Question 1 · Standard
  2. Question 2 · Basic
  3. Question 3 · Basic
  4. Question 4 · Basic
  5. Question 5 · Standard
  6. Question 6 · Standard

Using the quadratic formula

  1. Question 7 · Basic
  2. Question 8 · Basic
  3. Question 9 · Basic
  4. Question 10 · Basic
  5. Question 11 · Basic

Discriminant and nature of roots

  1. Question 12 · Basic
  2. Question 13 · Basic
  3. Question 14 · Basic
  4. Question 15 · Standard
  5. Question 16 · Basic

Unknown coefficients and equal roots

  1. Question 17 · Standard
  2. Question 18 · Standard
  3. Question 19 · Standard
  4. Question 20 · Standard
  5. Question 21 · Basic
  6. Question 22 · Basic
  7. Question 23 · Basic
  8. Question 24 · Basic
  9. Question 25 · Basic
  10. Question 26 · Standard
  11. Question 27 · Standard
  12. Question 28 · Basic
  13. Question 29 · Basic
  14. Question 30 · Basic
  15. Question 31 · Standard
  16. Question 32 · Standard
  17. Question 33 · Standard
  18. Question 34 · Standard

Forming equations and word problems

  1. Question 35 · Standard
  2. Question 36 · Standard
  3. Question 37 · Standard
  4. Question 38 · Standard

Solving by factorisation

Standard2 marks · Includes accessibqe-paper appearance

Q1. Find the roots of the quadratic equation 9x² − 6√2 x + 2 = 0. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

9x² − 3√2 x − 3√2 x + 2 = 0

3x(3x − √2) − √2(3x − √2) = 0

(3x − √2)² = 0

Both roots are √23

Basic2 marks

Q2. Solve for x: 2x² + 72x + 34 = 0. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

Multiply by 4: 8x² + 14x + 3 = 0.

8x² + 12x + 2x + 3 = 0

4x(2x + 3) + (2x + 3) = 0

(4x + 1)(2x + 3) = 0

x = −14 or x = −32

Basic2 marks · Includes accessibqe-paper appearance

Q3. Find the roots of the quadratic equation 3x² − 5x + 2 = 0. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

3x² − 3x − 2x + 2 = 0

3x(x − 1) − 2(x − 1) = 0

(3x − 2)(x − 1) = 0

3x − 2 = 0 or x − 1 = 0

x = 23 or x = 1

Basic2 marks

Q4. (a) Solve for y: 9y² − 64 = 0. [2 marks]

Show answer and solution

(3y)² − 8² = 0.

(3y − 8)(3y + 8) = 0.

y = 83 or y = −83

OR

(b) Write a quadratic equation whose roots are −7 and 9. [2 marks]

Appeared in: 2022 · Basic · Term II · Supplementary

Show answer and solution

Sum of roots = −7 + 9 = 2.

Product of roots = (−7)(9) = −63.

An equation is x² − (sum of roots)x + product of roots = 0.

One suitable equation is x² − 2x − 63 = 0

Standard2 marks

Q5. Solve the quadratic equation x² + 2√2 x − 6 = 0 for x. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

x² + 3√2 x − √2 x − 6 = 0

x(x + 3√2) − √2(x + 3√2) = 0

(x − √2)(x + 3√2) = 0

x = √2 or x = −3√2

Standard2 marks

Q6. Solve the quadratic equation x² − 2ax + (a² − b²) = 0 for x. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

x² − 2ax + a² − b² = 0

(x − a)² − b² = 0

(x − a − b)(x − a + b) = 0

x = a + b or x = a − b

Using the quadratic formula

Basic2 marks

Q7. Solve the quadratic equation 2x² − 5x − 1 = 0 for x. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

a = 2, b = −5, c = −1.

D = b² − 4ac = 25 + 8 = 33.

x = −b ± √D2a = 5 ± √334

x = 5 + √334 or x = 5 − √334

Basic2 marks

Q8. Solve the quadratic equation x² + 3x − 9 = 0 for x. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

a = 1, b = 3, c = −9.

D = b² − 4ac = 9 + 36 = 45.

x = −b ± √D2a = −3 ± √452

√45 = 3√5.

x = −3 + 3√52 or x = −3 − 3√52

Basic2 marks

Q9. Solve the equation 3x² − 8x − 1 = 0 for x. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

a = 3, b = −8, c = −1.

D = b² − 4ac = 64 + 12 = 76.

x = 8 ± √766

√76 = 2√19, so x = 4 ± √193.

x = 4 + √193 or x = 4 − √193

Basic2 marks

Q10. (a) Solve the equation 4x² − 9x + 3 = 0, using quadratic formula. [2 marks]

Show answer and solution

a = 4, b = −9, c = 3.

D = b² − 4ac = 81 − 48 = 33.

x = −b ± √D2a = 9 ± √338.

x = 9 + √338 or x = 9 − √338

OR

(b) Find the nature of roots of the equation 3x² − 4√3 x + 4 = 0. [2 marks]

Appeared in: 2025 · Basic · Main

Show answer and solution

a = 3, b = −4√3, c = 4.

D = b² − 4ac = 48 − 48 = 0.

Two equal real roots

Basic2 marks

Q11. (a) Solve the quadratic equation √3 x² + 10x + 7√3 = 0 using quadratic formula. [2 marks]

Show answer and solution

a = √3, b = 10, c = 7√3.

D = b² − 4ac = 100 − 84 = 16.

x = −10 ± 42√3.

x = −62√3 or −142√3.

x = −√3 or x = −7√33

OR

(b) Find the nature of roots of the equation 4x² − 4a²x + a⁴ − b⁴ = 0, b ≠ 0. [2 marks]

Appeared in: 2025 · Basic · Main

The first alternative also occurs in another OR pair on this page; the complete printed pairings are retained.

Show answer and solution

D = (−4a²)² − 4(4)(a⁴ − b⁴).

D = 16a⁴ − 16a⁴ + 16b⁴ = 16b⁴.

For real b ≠ 0, b⁴ > 0; hence D > 0.

Two distinct real roots

Discriminant and nature of roots

Basic2 marks

Q12. (a) Find the nature of the roots of the quadratic equation x² − 5x + 9 = 0. [2 marks]

Show answer and solution

a = 1, b = −5, c = 9.

D = b² − 4ac = 25 − 36 = −11.

D < 0.

No real roots

OR

(b) Write a quadratic equation with roots −3 and 5. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

An equation with roots α and β is x² − (α + β)x + αβ = 0.

α + β = −3 + 5 = 2; αβ = (−3)(5) = −15.

One suitable equation is x² − 2x − 15 = 0

Basic2 marks

Q13. Find the nature of the roots of the quadratic equation 4x² − 5x − 1 = 0. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

a = 4, b = −5, c = −1.

D = b² − 4ac = 25 + 16 = 41.

D > 0, so the equation has two distinct real roots.

Two distinct real roots (irrational)

Basic2 marks

Q14. (a) Find the discriminant of the quadratic equation 3x² − 2x + 13 = 0 and hence find the nature of its roots. [2 marks]

Show answer and solution

a = 3, b = −2, c = 13.

D = b² − 4ac = 4 − 4(3)(13) = 0.

Discriminant = 0; two equal real roots

OR

(b) Find the roots of the quadratic equation x² − x − 2 = 0. [2 marks]

Appeared in: 2023 · Basic · Main

Show answer and solution

x² − 2x + x − 2 = 0

x(x − 2) + (x − 2) = 0

(x − 2)(x + 1) = 0

x = 2 or x = −1

Standard2 marks

Q15. (a) Find the sum and product of the roots of the quadratic equation 2x² − 9x + 4 = 0. [2 marks]

Show answer and solution

For ax² + bx + c = 0, sum of roots = −b/a and product = c/a.

Here, a = 2, b = −9 and c = 4.

Sum = 92; product = 42 = 2.

Sum of roots = 92; product = 2

OR

(b) Find the discriminant of the quadratic equation 4x² − 5 = 0 and hence comment on the nature of roots of the equation. [2 marks]

Appeared in: 2023 · Standard · Main

Show answer and solution

a = 4, b = 0, c = −5.

D = b² − 4ac = 0 − 4(4)(−5) = 80.

D > 0.

Discriminant = 80; two distinct real roots

Basic2 marks

Q16. Find the nature of roots of the quadratic equation x² + 4x − 3√2 = 0. [2 marks]

Appeared in: 2025–26 · Basic · Official sample

Show answer and solution

a = 1, b = 4, c = −3√2.

D = b² − 4ac = 16 + 12√2.

Since √2 > 0, D > 0.

Two distinct real roots

Unknown coefficients and equal roots

Standard2 marks

Q17. (a) Find the value of m for which the quadratic equation (m − 1)x² + 2(m − 1)x + 1 = 0 has two real and equal roots. [2 marks]

Show answer and solution

Since the equation is quadratic, m − 1 ≠ 0.

D = [2(m − 1)]² − 4(m − 1)(1) = 0.

4(m − 1)(m − 2) = 0.

m = 1 or 2; reject m = 1 because it gives 1 = 0.

m = 2

OR

(b) Solve the following quadratic equation for x: √3 x² + 10x + 7√3 = 0. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

√3 x² + 3x + 7x + 7√3 = 0

√3 x(x + √3) + 7(x + √3) = 0

(√3 x + 7)(x + √3) = 0.

x = −7√33 or x = −√3

Standard2 marks

Q18. (a) Solve the quadratic equation for x: x² − 2ax − (4b² − a²) = 0. [2 marks]

Show answer and solution

x² − 2ax + a² − 4b² = 0.

(x − a)² − (2b)² = 0.

(x − a − 2b)(x − a + 2b) = 0.

x = a + 2b or x = a − 2b

OR

(b) If the quadratic equation (1 + a²)x² + 2abx + (b² − c²) = 0 has equal and real roots, then prove that b² = c²(1 + a²). [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

For real a, 1 + a² > 0; the equation is quadratic.

Equal roots give D = 0.

(2ab)² − 4(1 + a²)(b² − c²) = 0.

a²b² − (b² + a²b² − c² − a²c²) = 0.

−b² + c²(1 + a²) = 0.

Hence, b² = c²(1 + a²)

Standard2 marks

Q19. (a) If the sum of the roots of the quadratic equation ky² − 11y + (k − 23) = 0 is 1321 more than the product of the roots, then find the value of k. [2 marks]

Show answer and solution

Since the equation is quadratic, k ≠ 0.

Sum = 11k; product = k − 23k.

11k = k − 23k + 1321.

Multiply by 21k: 231 = 21k − 483 + 13k.

34k = 714.

k = 21

OR

(b) If x = −2 is the common solution of quadratic equations ax² + x − 3a = 0 and x² + bx + b = 0, then find the value of a²b. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

Substitute x = −2 in the first equation: 4a − 2 − 3a = 0, so a = 2.

In the second: 4 − 2b + b = 0, so b = 4.

a²b = 2² × 4.

a²b = 16

Standard2 marks

Q20. (a) Find the value of k for which the quadratic equation 2kx² − 40x + 25 = 0 has real and equal roots. [2 marks]

Show answer and solution

Since the equation is quadratic, k ≠ 0.

D = (−40)² − 4(2k)(25) = 0.

1600 − 200k = 0.

k = 8

OR

(b) Solve for x: 52x² + 25 = 1 − 2x. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

Multiply by 10 and rearrange: 25x² + 20x − 6 = 0.

D = 20² − 4(25)(−6) = 1000.

x = −20 ± √100050 = −20 ± 10√1050.

x = −2 + √105 or x = −2 − √105

Basic2 marks

Q21. (a) For what value of p does the quadratic equation px² + 2x + p = 0 have real and equal roots? [2 marks]

Show answer and solution

Since the equation is quadratic, p ≠ 0.

D = 2² − 4(p)(p) = 0.

4 − 4p² = 0, so p² = 1.

p = 1 or p = −1

OR

(b) Solve the quadratic equation for x: 6 − x − x² = 0. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

Rearrange: x² + x − 6 = 0.

x² + 3x − 2x − 6 = 0.

x(x + 3) − 2(x + 3) = 0.

(x − 2)(x + 3) = 0.

x = 2 or x = −3

Basic2 marks

Q22. Find the value of k so that the quadratic equation 3x² − 5x − 2k = 0 has real and equal roots. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

For real and equal roots, D = b² − 4ac = 0.

25 − 4(3)(−2k) = 0.

25 + 24k = 0.

k = −2524

Basic2 marks

Q23. Find the value(s) of a for which the quadratic equation x² − ax + 1 = 0 has real and equal roots. [2 marks]

Appeared in: 2022 · Basic · Term II · Main

Show answer and solution

For real and equal roots, D = 0.

(−a)² − 4(1)(1) = 0.

a² − 4 = 0.

(a − 2)(a + 2) = 0.

a = 2 or a = −2

Basic2 marks

Q24. (a) Solve for x: 6x² − 2x + 16 = 0. [2 marks]

Show answer and solution

Multiply by 6: 36x² − 12x + 1 = 0.

(6x − 1)² = 0.

6x − 1 = 0.

Both roots are 16

OR

(b) Find the value of k for which the quadratic equation 9x² + 8kx + 16 = 0 has real and equal roots. [2 marks]

Appeared in: 2022 · Basic · Term II · Supplementary

Show answer and solution

For real and equal roots, D = 0.

(8k)² − 4(9)(16) = 0.

64k² = 576, so k² = 9.

k = 3 or k = −3

Basic2 marks · Includes accessibqe-paper appearance

Q25. Find the value of k for the quadratic equation kx² − 4kx + 12 = 0 to have two real and equal roots. [2 marks]

Appeared in: 2022 · Basic · Term II · Supplementary

Show answer and solution

Since the equation is quadratic, k ≠ 0.

D = (−4k)² − 4(k)(12) = 0.

16k(k − 3) = 0, so k = 0 or k = 3.

Reject k = 0: the equation would become 12 = 0.

k = 3

Standard2 marks

Q26. (a) Solve for x: 2x² − 2√2 x + 1 = 0. [2 marks]

Show answer and solution

2x² − 2√2 x + 1 = 0.

(√2 x − 1)² = 0.

√2 x = 1.

Both roots are √22

OR

(b) Find the value(s) of k for which the quadratic equation x² + 5kx + 16 = 0 has real and equal roots. [2 marks]

Appeared in: 2022 · Standard · Term II · Supplementary

Show answer and solution

For real and equal roots, D = 0.

(5k)² − 4(1)(16) = 0.

25k² = 64.

k = 85 or k = −85

Standard2 marks

Q27. (a) Find the value of k for which the quadratic equation kx² − 5x + k = 0 has real and equal roots. [2 marks]

Show answer and solution

Since the equation is quadratic, k ≠ 0.

D = (−5)² − 4(k)(k) = 0.

4k² = 25.

k = 52 or k = −52

OR

(b) Solve for y: y² + 3√52y − 5 = 0. [2 marks]

Appeared in: 2022 · Standard · Term II · Supplementary

Show answer and solution

y² + 2√5 y − √52y − 5 = 0.

y(y + 2√5) − √52(y + 2√5) = 0.

(y − √52)(y + 2√5) = 0.

y = √52 or y = −2√5

Basic2 marks

Q28. (a) Find the value of k for which the roots of the quadratic equation 5x² − 10x + k = 0 are real and equal. [2 marks]

Show answer and solution

For real and equal roots, D = 0.

(−10)² − 4(5)(k) = 0.

100 − 20k = 0.

k = 5

OR

(b) If one root of the quadratic equation 3x² − 8x − (2k + 1) = 0 is seven times the other, then find the value of k. [2 marks]

Appeared in: 2023 · Basic · Main

Show answer and solution

Let the roots be t and 7t.

t + 7t = 83, so t = 13.

Their product is 7t² = 79.

−(2k + 1)3 = 79, so −6k − 3 = 7.

k = −53

Basic2 marks

Q29. If 23 is a root of the quadratic equation kx² − x − 2 = 0, then find the value of k. [2 marks]

Appeared in: 2024 · Basic · Main

Show answer and solution

Substitute x = 23: k(23)² − 23 − 2 = 0.

4k9 = 83.

k = 83 × 94 = 6.

k = 6

Basic2 marks

Q30. If 12 is a root of the quadratic equation x² + kx − 54 = 0, then find the value of k. [2 marks]

Appeared in: 2024 · Basic · Main

Show answer and solution

Substitute x = 12: 14 + k2 − 54 = 0.

k2 = 1.

k = 2

Standard2 marks

Q31. (a) Find the value(s) of k so that the quadratic equation 4x² + kx + 1 = 0 has real and equal roots. [2 marks]

Show answer and solution

For real and equal roots, D = 0.

k² − 4(4)(1) = 0.

k² = 16.

k = 4 or k = −4

OR

(b) If α and β are the zeroes of the polynomial p(y) = y² − 5y + 3, then find the value of α⁴β³ + α³β⁴. [2 marks]

Appeared in: 2025 · Standard · Main

Show answer and solution

α + β = 5; αβ = 3.

α⁴β³ + α³β⁴ = α³β³(α + β).

= (αβ)³(α + β) = 3³ × 5.

α⁴β³ + α³β⁴ = 135

Standard2 marks

Q32. Verify that roots of the quadratic equation (p − q)x² + (q − r)x + (r − p) = 0 are equal when q + r = 2p. [2 marks]

Appeared in: 2026 · Standard · Main

The printed question calls the equation quadratic, so p − q ≠ 0 is required.

Show answer and solution

Since the question states a quadratic equation, p − q ≠ 0.

q + r = 2p gives r = 2p − q.

Substitute: (p − q)x² − 2(p − q)x + (p − q) = 0.

Divide by p − q: x² − 2x + 1 = 0.

(x − 1)² = 0.

The roots are equal: 1 and 1

Standard2 marks

Q33. If α and β are the roots of quadratic equation x² + (k² − 1)x − 20 = 0 such that α² − β² − αβ = 29 and α − β = 9, then find the value of k. [2 marks]

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

α + β = 1 − k²; αβ = −20.

α² − β² − αβ = (α − β)(α + β) − αβ.

9(1 − k²) + 20 = 29.

29 − 9k² = 29, so k² = 0.

k = 0

Standard2 marks · Includes accessibqe-paper appearance

Q34. If 4 is a root of the equation x² + kx + 8 = 0 then find the other root. Also find the value of k. [2 marks]

Appeared in: 2022 · Standard · Term II · Supplementary

Show answer and solution

Let the other root be β. Product of roots = ca = 8.

4β = 8, so β = 2.

Sum of roots = −ba = −k.

4 + 2 = −k.

Other root = 2; k = −6

Forming equations and word problems

Standard2 marks

Q35. The product of Rehan’s age (in years) 5 years ago and his age 7 years from now is one more than twice his present age. Find his present age. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

Let Rehan’s present age be x years; x > 5.

(x − 5)(x + 7) = 2x + 1.

x² + 2x − 35 = 2x + 1, so x² − 36 = 0.

(x − 6)(x + 6) = 0, so x = 6 or −6.

Reject −6 because age cannot be negative.

Rehan’s present age is 6 years

Standard2 marks

Q36. (a) For what value of m does the quadratic equation mx² − 2(m − 1)x + (m + 2) = 0 have two real and equal roots? [2 marks]

Show answer and solution

Since the equation is quadratic, m ≠ 0.

D = [−2(m − 1)]² − 4m(m + 2) = 0.

4[(m − 1)² − m(m + 2)] = 0.

1 − 4m = 0.

m = 14

OR

(b) The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

Let the shorter side be x m; the longer side is (x + 30) m and diagonal (x + 60) m.

By Pythagoras, x² + (x + 30)² = (x + 60)².

x² − 60x − 2700 = 0.

(x − 90)(x + 30) = 0, so x = 90 or −30.

Reject −30; the longer side is 90 + 30 = 120 m.

The sides are 90 m and 120 m

Standard2 marks · Includes accessibqe-paper appearance

Q37. (a) The product of two consecutive odd positive integers is 255. Find the integers by formulating a quadratic equation. [2 marks]

Show answer and solution

Let the integers be x and x + 2, where x is positive and odd.

x(x + 2) = 255, so x² + 2x − 255 = 0.

(x − 15)(x + 17) = 0.

x = 15 or −17; reject −17 because the integers are positive.

The integers are 15 and 17

OR

(b) Find the value(s) of k for the quadratic equation (k + 3)x² + kx + 1 = 0 to have two real and equal roots. [2 marks]

Appeared in: 2022 · Standard · Term II · Main

Show answer and solution

Since the equation is quadratic, k ≠ −3.

D = k² − 4(k + 3) = 0.

k² − 4k − 12 = 0.

(k − 6)(k + 2) = 0; both values keep k + 3 ≠ 0.

k = 6 or k = −2

Standard2 marks

Q38. The area of a right-angled triangle is 600 sq. cm. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle. [2 marks]

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

Let the altitude be x cm and base (x + 10) cm, with x > 0.

12x(x + 10) = 600.

x² + 10x − 1200 = 0.

(x − 30)(x + 40) = 0, so x = 30 or −40.

Reject −40; the base is 30 + 10 = 40 cm.

Hypotenuse = √(30² + 40²) = 50 cm.

Altitude = 30 cm; base = 40 cm; hypotenuse = 50 cm

Questions students ask

How do I identify a quadratic equation?

Simplify the equation and bring every term to one side. It is quadratic when the highest remaining power of the unknown is 2 and its coefficient is non-zero. An equation that loses its x² term after simplification is not quadratic.

When should I use the quadratic formula?

Use it when the question asks for it or when factorisation is inconvenient. Write the equation as ax² + bx + c = 0, identify the coefficients, calculate D = b² − 4ac, then use x = (−b ± √D)/(2a) for real roots when D ≥ 0.

Do I need to find the roots to state their nature?

No. Calculate the discriminant D. A positive D gives two distinct real roots, zero gives two equal real roots, and a negative D gives no real roots. The sign alone is enough unless the question also asks for the roots.

Why must I check the coefficient of x² in a parameter question?

The condition D = 0 applies to a quadratic equation, so its leading coefficient must be non-zero. A candidate parameter value may remove the x² term or even produce a contradiction. Substitute each candidate into the original equation before accepting it.

Can both roots be correct in a word problem?

Yes, if both satisfy every condition in the passage. Check positivity, units, integer requirements and any stated restrictions. Reject a root only with a reason; a negative root is valid in a pure algebra question unless the question restricts it.

How do I avoid sign mistakes in the discriminant?

Read a, b and c from the standard form, including their signs. Put negative coefficients in brackets when substituting. For example, b = −5 gives b² = (−5)² = 25, and a negative c makes −4ac positive when a is positive.

Is the quadratic equation formed from two roots unique?

No. Multiplying a quadratic equation by any non-zero constant keeps its roots. The monic equation x² − (α + β)x + αβ = 0 is one suitable choice when the roots are α and β.

Do I solve both OR alternatives in the exam?

Answer one alternative where an OR choice is printed. Both alternatives are solved here for practice. Each alternative carries 2 marks, but the complete OR pair counts as one numbered question on this page.

Answer summary

Open the answer summary
QuestionAnswer
Q1Both roots are √23
Q2x = −14 or x = −32
Q3x = 23 or x = 1
Q4(a)y = 83 or y = −83
Q4(b) OROne suitable equation is x² − 2x − 63 = 0
Q5x = √2 or x = −3√2
Q6x = a + b or x = a − b
Q7x = 5 + √334 or x = 5 − √334
Q8x = −3 + 3√52 or x = −3 − 3√52
Q9x = 4 + √193 or x = 4 − √193
Q10(a)x = 9 + √338 or x = 9 − √338
Q10(b) ORTwo equal real roots
Q11(a)x = −√3 or x = −7√33
Q11(b) ORTwo distinct real roots
Q12(a)No real roots
Q12(b) OROne suitable equation is x² − 2x − 15 = 0
Q13Two distinct real roots (irrational)
Q14(a)Discriminant = 0; two equal real roots
Q14(b) ORx = 2 or x = −1
Q15(a)Sum of roots = 92; product = 2
Q15(b) ORDiscriminant = 80; two distinct real roots
Q16Two distinct real roots
Q17(a)m = 2
Q17(b) ORx = −7√33 or x = −√3
Q18(a)x = a + 2b or x = a − 2b
Q18(b) ORHence, b² = c²(1 + a²)
Q19(a)k = 21
Q19(b) ORa²b = 16
Q20(a)k = 8
Q20(b) ORx = −2 + √105 or x = −2 − √105
Q21(a)p = 1 or p = −1
Q21(b) ORx = 2 or x = −3
Q22k = −2524
Q23a = 2 or a = −2
Q24(a)Both roots are 16
Q24(b) ORk = 3 or k = −3
Q25k = 3
Q26(a)Both roots are √22
Q26(b) ORk = 85 or k = −85
Q27(a)k = 52 or k = −52
Q27(b) ORy = √52 or y = −2√5
Q28(a)k = 5
Q28(b) ORk = −53
Q29k = 6
Q30k = 2
Q31(a)k = 4 or k = −4
Q31(b) ORα⁴β³ + α³β⁴ = 135
Q32The roots are equal: 1 and 1
Q33k = 0
Q34Other root = 2; k = −6
Q35Rehan’s present age is 6 years
Q36(a)m = 14
Q36(b) ORThe sides are 90 m and 120 m
Q37(a)The integers are 15 and 17
Q37(b) ORk = 6 or k = −2
Q38Altitude = 30 cm; base = 40 cm; hypotenuse = 50 cm

Both OR alternatives are listed for practice. The answer summary follows your selected filters.

How to approach a 2-mark question

  1. Identify what is required and follow any specified method.
  2. Write the quadratic equation in standard form; identify coefficients or define the unknown with units.
  3. Show one logical operation per line, keeping signs and brackets.
  4. State both roots or the nature of roots; check the original equation and any contextual restrictions.

About these questions and solutions

Selected official CBSE 2-mark questions; exact repeated questions are merged, and complete OR pairings are retained.
Solutions prepared by Team TutorVertex in NCERT style, with answers checked against the cited marking schemes.

Official sources — question papers and marking schemes

All page references below are PDF pages. Archive member names identify the exact files inside official ZIPs. Question wording follows the first listed appearance; spacing and punctuation are normalised.

Question 1 sources

  • 2022 · Standard · Term II · Main · 30/B/5 · Q2Official question paper · PDF p. 3
    Math_S/30-B-5 Maths Std. for VI candidates.pdf
    Official marking scheme · PDF pp. 3, 4
    Mathematic_Standard/041_30_B_5_Math (Standard) (For VI).pdf

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Question 11 sources

  • 2025 · Basic · Main · 430/6/3 · Q23Official question paper · PDF p. 13
    241_Mathematics_Basic/430-6-3_Mathematics Basic.pdf
    Official marking scheme · PDF pp. 8, 9
    Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_6_3.pdf

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Question 31 sources

  • 2025 · Standard · Main · 30/3/1 · Q24Official question paper · PDF p. 13
    041_Mathematics_Standard/30-3-1_Mathematics Standard.pdf
    Official marking scheme · PDF p. 9
    Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-3-1_2024-25.pdf
  • 2025 · Standard · Main · 30/3/3 · Q22Official question paper · PDF p. 13
    041_Mathematics_Standard/30-3-3_Mathematics Standard.pdf
    Official marking scheme · PDF p. 8
    Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-3-3_2024-25.pdf

Question 32 sources

  • 2026 · Standard · Main · 30/4/1 · Q21Official question paper · PDF p. 13
    Mathematics_Standard/1171-1_30-4-1 (Mathematics Standard).pdf
    Official marking scheme · PDF p. 8
    041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-4-1_2025-26 (1).pdf
  • 2026 · Standard · Main · 30/4/2 · Q23Official question paper · PDF p. 13
    Mathematics_Standard/1171-2_30-4-2 (Mathematics Standard).pdf
    Official marking scheme · PDF p. 9
    041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-4-2_2025-26 (1).pdf
  • 2026 · Standard · Main · 30/4/3 · Q24Official question paper · PDF p. 13
    Mathematics_Standard/1171-3_30-4-3 (Mathematics Standard).pdf
    Official marking scheme · PDF p. 10
    041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-4-3_2025-26 (1).pdf

Question 33 sources

  • 2026 · Standard · Second board examination · 30/8/1 · Q22Official question paper · PDF p. 17
    Mathmatics_Std/30-8-1_Maths Std_E2_10_2026.pdf
    Official marking scheme · PDF p. 8
    Mathematics_Standard/MS_X_041_Mathematics_30-8-1_2025-26 F (1).pdf
  • 2026 · Standard · Second board examination · 30/8/2 · Q23Official question paper · PDF p. 17
    Mathmatics_Std/30-8-2_Maths Std_E2_10_2026.pdf
    Official marking scheme · PDF p. 9
    Mathematics_Standard/MS_X_041_Mathematics_30-8-2_2025-26 F (1).pdf

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Question 38 sources

  • 2026 · Standard · Second board examination · 30/8/3 · Q24Official question paper · PDF p. 17
    Mathmatics_Std/30-8-3_Maths Std_E2_10_2026.pdf
    Official marking scheme · PDF pp. 9, 10
    Mathematics_Standard/MS_X_041_Mathematics_30-8-3_2025-26 F (1).pdf

NCERT Class 10, Chapter 4: Quadratic Equations
CBSE Mathematics syllabus, 2026–27

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