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
Chapter 4 · Written-answer practice
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
| Use | Result or method |
|---|---|
| Standard form | ax² + bx + c = 0, a ≠ 0. Here a, b and c are the coefficients after simplification. |
| Discriminant | D = b² − 4ac. |
| Nature of real roots | D > 0: two distinct real roots; D = 0: equal real roots; D < 0: no real roots. |
| Quadratic formula | x = , for D ≥ 0. |
| Root sum and product | α + β = ; αβ = . |
| Zero-product rule | If 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.
| Question | A useful starting point |
|---|---|
| Roots by factorisation | Bring all terms to one side, split the middle term or use a suitable identity, then set each factor equal to zero. |
| Quadratic formula is requested | Identify a, b and c; calculate D; substitute in the formula and simplify both roots. |
| Only the nature of roots | Calculate D and state the conclusion from its sign. |
| Real and equal roots | Set D = 0, solve for the parameter and check a ≠ 0. |
| A root is given | Substitute it into the original equation; use the root sum or product if another root is needed. |
| A short word problem | Define 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.
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Q1. Find the roots of the quadratic equation 9x² − 6√2 x + 2 = 0. [2 marks]
Appeared in: 2022 · Standard · Term II · Main
9x² − 3√2 x − 3√2 x + 2 = 0
3x(3x − √2) − √2(3x − √2) = 0
(3x − √2)² = 0
Both roots are
Q2. Solve for x: 2x² + x + = 0. [2 marks]
Appeared in: 2022 · Basic · Term II · Main
Multiply by 4: 8x² + 14x + 3 = 0.
8x² + 12x + 2x + 3 = 0
4x(2x + 3) + (2x + 3) = 0
(4x + 1)(2x + 3) = 0
x = or x =
Q3. Find the roots of the quadratic equation 3x² − 5x + 2 = 0. [2 marks]
Appeared in: 2022 · Basic · Term II · Main
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 = or x = 1
Q4. (a) Solve for y: 9y² − 64 = 0. [2 marks]
(3y)² − 8² = 0.
(3y − 8)(3y + 8) = 0.
y = or y =
OR
(b) Write a quadratic equation whose roots are −7 and 9. [2 marks]
Appeared in: 2022 · Basic · Term II · Supplementary
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
Q5. Solve the quadratic equation x² + 2√2 x − 6 = 0 for x. [2 marks]
Appeared in: 2022 · Standard · Term II · Main
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
Q6. Solve the quadratic equation x² − 2ax + (a² − b²) = 0 for x. [2 marks]
Appeared in: 2022 · Standard · Term II · Main
x² − 2ax + a² − b² = 0
(x − a)² − b² = 0
(x − a − b)(x − a + b) = 0
x = a + b or x = a − b
Q7. Solve the quadratic equation 2x² − 5x − 1 = 0 for x. [2 marks]
Appeared in: 2022 · Basic · Term II · Main
a = 2, b = −5, c = −1.
D = b² − 4ac = 25 + 8 = 33.
x = =
x = or x =
Q8. Solve the quadratic equation x² + 3x − 9 = 0 for x. [2 marks]
Appeared in: 2022 · Basic · Term II · Main
a = 1, b = 3, c = −9.
D = b² − 4ac = 9 + 36 = 45.
x = =
√45 = 3√5.
x = or x =
Q9. Solve the equation 3x² − 8x − 1 = 0 for x. [2 marks]
Appeared in: 2022 · Basic · Term II · Main
a = 3, b = −8, c = −1.
D = b² − 4ac = 64 + 12 = 76.
x =
√76 = 2√19, so x = .
x = or x =
Q10. (a) Solve the equation 4x² − 9x + 3 = 0, using quadratic formula. [2 marks]
a = 4, b = −9, c = 3.
D = b² − 4ac = 81 − 48 = 33.
x = = .
x = or x =
OR
(b) Find the nature of roots of the equation 3x² − 4√3 x + 4 = 0. [2 marks]
Appeared in: 2025 · Basic · Main
a = 3, b = −4√3, c = 4.
D = b² − 4ac = 48 − 48 = 0.
Two equal real roots
Q11. (a) Solve the quadratic equation √3 x² + 10x + 7√3 = 0 using quadratic formula. [2 marks]
a = √3, b = 10, c = 7√3.
D = b² − 4ac = 100 − 84 = 16.
x = .
x = or .
x = −√3 or x =
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.
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
Q12. (a) Find the nature of the roots of the quadratic equation x² − 5x + 9 = 0. [2 marks]
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
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
Q13. Find the nature of the roots of the quadratic equation 4x² − 5x − 1 = 0. [2 marks]
Appeared in: 2022 · Basic · Term II · Main
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)
Q14. (a) Find the discriminant of the quadratic equation 3x² − 2x + = 0 and hence find the nature of its roots. [2 marks]
a = 3, b = −2, c = .
D = b² − 4ac = 4 − 4(3)() = 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
x² − 2x + x − 2 = 0
x(x − 2) + (x − 2) = 0
(x − 2)(x + 1) = 0
x = 2 or x = −1
Q15. (a) Find the sum and product of the roots of the quadratic equation 2x² − 9x + 4 = 0. [2 marks]
For ax² + bx + c = 0, sum of roots = −b/a and product = c/a.
Here, a = 2, b = −9 and c = 4.
Sum = ; product = = 2.
Sum of roots = ; 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
a = 4, b = 0, c = −5.
D = b² − 4ac = 0 − 4(4)(−5) = 80.
D > 0.
Discriminant = 80; two distinct real roots
Q16. Find the nature of roots of the quadratic equation x² + 4x − 3√2 = 0. [2 marks]
Appeared in: 2025–26 · Basic · Official sample
a = 1, b = 4, c = −3√2.
D = b² − 4ac = 16 + 12√2.
Since √2 > 0, D > 0.
Two distinct real roots
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]
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
√3 x² + 3x + 7x + 7√3 = 0
√3 x(x + √3) + 7(x + √3) = 0
(√3 x + 7)(x + √3) = 0.
x = or x = −√3
Q18. (a) Solve the quadratic equation for x: x² − 2ax − (4b² − a²) = 0. [2 marks]
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
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)
Q19. (a) If the sum of the roots of the quadratic equation ky² − 11y + (k − 23) = 0 is more than the product of the roots, then find the value of k. [2 marks]
Since the equation is quadratic, k ≠ 0.
Sum = ; product = .
= + .
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
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
Q20. (a) Find the value of k for which the quadratic equation 2kx² − 40x + 25 = 0 has real and equal roots. [2 marks]
Since the equation is quadratic, k ≠ 0.
D = (−40)² − 4(2k)(25) = 0.
1600 − 200k = 0.
k = 8
OR
(b) Solve for x: x² + = 1 − 2x. [2 marks]
Appeared in: 2022 · Standard · Term II · Main
Multiply by 10 and rearrange: 25x² + 20x − 6 = 0.
D = 20² − 4(25)(−6) = 1000.
x = = .
x = or x =
Q21. (a) For what value of p does the quadratic equation px² + 2x + p = 0 have real and equal roots? [2 marks]
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
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
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
For real and equal roots, D = b² − 4ac = 0.
25 − 4(3)(−2k) = 0.
25 + 24k = 0.
k =
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
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
Q24. (a) Solve for x: 6x² − 2x + = 0. [2 marks]
Multiply by 6: 36x² − 12x + 1 = 0.
(6x − 1)² = 0.
6x − 1 = 0.
Both roots are
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
For real and equal roots, D = 0.
(8k)² − 4(9)(16) = 0.
64k² = 576, so k² = 9.
k = 3 or k = −3
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
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
Q26. (a) Solve for x: 2x² − 2√2 x + 1 = 0. [2 marks]
2x² − 2√2 x + 1 = 0.
(√2 x − 1)² = 0.
√2 x = 1.
Both roots are
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
For real and equal roots, D = 0.
(5k)² − 4(1)(16) = 0.
25k² = 64.
k = or k =
Q27. (a) Find the value of k for which the quadratic equation kx² − 5x + k = 0 has real and equal roots. [2 marks]
Since the equation is quadratic, k ≠ 0.
D = (−5)² − 4(k)(k) = 0.
4k² = 25.
k = or k =
OR
(b) Solve for y: y² + y − 5 = 0. [2 marks]
Appeared in: 2022 · Standard · Term II · Supplementary
y² + 2√5 y − y − 5 = 0.
y(y + 2√5) − (y + 2√5) = 0.
(y − )(y + 2√5) = 0.
y = or y = −2√5
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]
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
Let the roots be t and 7t.
t + 7t = , so t = .
Their product is 7t² = .
= , so −6k − 3 = 7.
k =
Q29. If is a root of the quadratic equation kx² − x − 2 = 0, then find the value of k. [2 marks]
Appeared in: 2024 · Basic · Main
Substitute x = : k()² − − 2 = 0.
= .
k = × = 6.
k = 6
Q30. If is a root of the quadratic equation x² + kx − = 0, then find the value of k. [2 marks]
Appeared in: 2024 · Basic · Main
Substitute x = : + − = 0.
= 1.
k = 2
Q31. (a) Find the value(s) of k so that the quadratic equation 4x² + kx + 1 = 0 has real and equal roots. [2 marks]
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
α + β = 5; αβ = 3.
α⁴β³ + α³β⁴ = α³β³(α + β).
= (αβ)³(α + β) = 3³ × 5.
α⁴β³ + α³β⁴ = 135
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.
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
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
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
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
Let the other root be β. Product of roots = = 8.
4β = 8, so β = 2.
Sum of roots = = −k.
4 + 2 = −k.
Other root = 2; k = −6
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
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
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]
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 =
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
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
Q37. (a) The product of two consecutive odd positive integers is 255. Find the integers by formulating a quadratic equation. [2 marks]
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
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
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
Let the altitude be x cm and base (x + 10) cm, with x > 0.
x(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
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.
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.
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.
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.
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.
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.
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.
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.
| Question | Answer |
|---|---|
| Q1 | Both roots are |
| Q2 | x = or x = |
| Q3 | x = or x = 1 |
| Q4(a) | y = or y = |
| Q4(b) OR | One suitable equation is x² − 2x − 63 = 0 |
| Q5 | x = √2 or x = −3√2 |
| Q6 | x = a + b or x = a − b |
| Q7 | x = or x = |
| Q8 | x = or x = |
| Q9 | x = or x = |
| Q10(a) | x = or x = |
| Q10(b) OR | Two equal real roots |
| Q11(a) | x = −√3 or x = |
| Q11(b) OR | Two distinct real roots |
| Q12(a) | No real roots |
| Q12(b) OR | One suitable equation is x² − 2x − 15 = 0 |
| Q13 | Two distinct real roots (irrational) |
| Q14(a) | Discriminant = 0; two equal real roots |
| Q14(b) OR | x = 2 or x = −1 |
| Q15(a) | Sum of roots = ; product = 2 |
| Q15(b) OR | Discriminant = 80; two distinct real roots |
| Q16 | Two distinct real roots |
| Q17(a) | m = 2 |
| Q17(b) OR | x = or x = −√3 |
| Q18(a) | x = a + 2b or x = a − 2b |
| Q18(b) OR | Hence, b² = c²(1 + a²) |
| Q19(a) | k = 21 |
| Q19(b) OR | a²b = 16 |
| Q20(a) | k = 8 |
| Q20(b) OR | x = or x = |
| Q21(a) | p = 1 or p = −1 |
| Q21(b) OR | x = 2 or x = −3 |
| Q22 | k = |
| Q23 | a = 2 or a = −2 |
| Q24(a) | Both roots are |
| Q24(b) OR | k = 3 or k = −3 |
| Q25 | k = 3 |
| Q26(a) | Both roots are |
| Q26(b) OR | k = or k = |
| Q27(a) | k = or k = |
| Q27(b) OR | y = or y = −2√5 |
| Q28(a) | k = 5 |
| Q28(b) OR | k = |
| Q29 | k = 6 |
| Q30 | k = 2 |
| Q31(a) | k = 4 or k = −4 |
| Q31(b) OR | α⁴β³ + α³β⁴ = 135 |
| Q32 | The roots are equal: 1 and 1 |
| Q33 | k = 0 |
| Q34 | Other root = 2; k = −6 |
| Q35 | Rehan’s present age is 6 years |
| Q36(a) | m = |
| Q36(b) OR | The sides are 90 m and 120 m |
| Q37(a) | The integers are 15 and 17 |
| Q37(b) OR | k = 6 or k = −2 |
| Q38 | Altitude = 30 cm; base = 40 cm; hypotenuse = 50 cm |
Both OR alternatives are listed for practice. The answer summary follows your selected filters.
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.
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.
NCERT Class 10, Chapter 4: Quadratic Equations
CBSE Mathematics syllabus, 2026–27
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