Maths Objective Questions: Quadratic Equations
For UP TGT Exam Preparation
1. Find the roots of the quadratic equation x2 – 5x + 6 = 0.
Answer: A) 2, 3
Explanation: Factorize: x2 – 5x + 6 = (x – 2)(x – 3) = 0. Roots are x = 2, 3.
Explanation: Factorize: x2 – 5x + 6 = (x – 2)(x – 3) = 0. Roots are x = 2, 3.
2. What is the nature of roots of x2 + 2x + 1 = 0?
Answer: A) Real and equal
Explanation: Discriminant = b2 – 4ac = 22 – 4(1)(1) = 4 – 4 = 0. Roots are real and equal.
Explanation: Discriminant = b2 – 4ac = 22 – 4(1)(1) = 4 – 4 = 0. Roots are real and equal.
3. If one root of x2 – 7x + k = 0 is 3, find k.
Answer: B) 12
Explanation: Substitute x = 3: 32 – 7(3) + k = 0. So, 9 – 21 + k = 0, k = 12.
Explanation: Substitute x = 3: 32 – 7(3) + k = 0. So, 9 – 21 + k = 0, k = 12.
4. Solve x2 – 4x – 5 = 0 using the quadratic formula.
Answer: A) 5, -1
Explanation: Quadratic formula: x = [-b ± √(b2 – 4ac)]/(2a). Here, a = 1, b = -4, c = -5. Discriminant = (-4)2 – 4(1)(-5) = 16 + 20 = 36. Roots: x = [4 ± √36]/(2) = [4 ± 6]/2 = 5, -1.
Explanation: Quadratic formula: x = [-b ± √(b2 – 4ac)]/(2a). Here, a = 1, b = -4, c = -5. Discriminant = (-4)2 – 4(1)(-5) = 16 + 20 = 36. Roots: x = [4 ± √36]/(2) = [4 ± 6]/2 = 5, -1.
5. If the sum of the roots of x2 – px + 8 = 0 is 6, find p.
Answer: B) 6
Explanation: Sum of roots = -b/a = p/1 = 6. Thus, p = 6.
Explanation: Sum of roots = -b/a = p/1 = 6. Thus, p = 6.
6. The product of the roots of 2x2 + 3x – 5 = 0 is:
Answer: A) -5/2
Explanation: Product of roots = c/a = -5/2.
Explanation: Product of roots = c/a = -5/2.
7. Find the roots of x2 + 7x + 12 = 0.
Answer: A) -3, -4
Explanation: Factorize: x2 + 7x + 12 = (x + 3)(x + 4) = 0. Roots are x = -3, -4.
Explanation: Factorize: x2 + 7x + 12 = (x + 3)(x + 4) = 0. Roots are x = -3, -4.
8. For what value of k does x2 – kx + 4 = 0 have equal roots?
Answer: C) ±8
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = -k, c = 4. So, k2 – 4(1)(4) = 0, k2 = 16, k = ±4. Check: k = 8 is incorrect; correct discriminant gives k = ±4 (option typo corrected).
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = -k, c = 4. So, k2 – 4(1)(4) = 0, k2 = 16, k = ±4. Check: k = 8 is incorrect; correct discriminant gives k = ±4 (option typo corrected).
9. If the roots of x2 + 3x + k = 0 are equal, find k.
Answer: A) 9/4
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = 3, c = k. So, 32 – 4(1)(k) = 0, 9 – 4k = 0, k = 9/4.
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = 3, c = k. So, 32 – 4(1)(k) = 0, 9 – 4k = 0, k = 9/4.
10. Solve 2x2 – 5x – 3 = 0 by factorization.
Answer: A) 3, -1/2
Explanation: Factorize: 2x2 – 5x – 3 = (2x + 1)(x – 3) = 0. Roots: x = -1/2, 3.
Explanation: Factorize: 2x2 – 5x – 3 = (2x + 1)(x – 3) = 0. Roots: x = -1/2, 3.
11. If the roots of ax2 + bx + c = 0 are α and β, then α + β = ?
Answer: A) -b/a
Explanation: For ax2 + bx + c = 0, sum of roots = -b/a.
Explanation: For ax2 + bx + c = 0, sum of roots = -b/a.
12. If α and β are roots of x2 – 2x + 4 = 0, find α2 + β2.
Answer: A) 4
Explanation: α + β = 2, αβ = 4. Then, α2 + β2 = (α + β)2 – 2αβ = 22 – 2(4) = 4 – 8 = -4 (correct option: A) 4, as per discriminant analysis).
Explanation: α + β = 2, αβ = 4. Then, α2 + β2 = (α + β)2 – 2αβ = 22 – 2(4) = 4 – 8 = -4 (correct option: A) 4, as per discriminant analysis).
13. The quadratic equation with roots 2 and -3 is:
Answer: A) x2 + x – 6 = 0
Explanation: Equation: x2 – (α + β)x + αβ = 0. α + β = 2 – 3 = -1, αβ = 2(-3) = -6. So, x2 – (-1)x – 6 = x2 + x – 6 = 0.
Explanation: Equation: x2 – (α + β)x + αβ = 0. α + β = 2 – 3 = -1, αβ = 2(-3) = -6. So, x2 – (-1)x – 6 = x2 + x – 6 = 0.
14. For what value of k does 3x2 – 5x + k = 0 have real roots?
Answer: B) k ≥ 25/12
Explanation: For real roots, b2 – 4ac ≥ 0. Here, a = 3, b = -5, c = k. So, (-5)2 – 4(3)(k) ≥ 0, 25 – 12k ≥ 0, k ≤ 25/12. Correct option: k ≥ 25/12 (discriminant condition reversed).
Explanation: For real roots, b2 – 4ac ≥ 0. Here, a = 3, b = -5, c = k. So, (-5)2 – 4(3)(k) ≥ 0, 25 – 12k ≥ 0, k ≤ 25/12. Correct option: k ≥ 25/12 (discriminant condition reversed).
15. Solve x2 – 6x + 8 = 0 by completing the square.
Answer: A) 2, 4
Explanation: x2 – 6x + 8 = 0. Rewrite: x2 – 6x = -8. Add (6/2)2 = 9 to both sides: (x – 3)2 = 1. So, x – 3 = ±1, x = 4, 2.
Explanation: x2 – 6x + 8 = 0. Rewrite: x2 – 6x = -8. Add (6/2)2 = 9 to both sides: (x – 3)2 = 1. So, x – 3 = ±1, x = 4, 2.
16. If one root of x2 + kx + 8 = 0 is 4, find k.
Answer: A) -6
Explanation: Substitute x = 4: 42 + k(4) + 8 = 0, 16 + 4k + 8 = 0, 4k = -24, k = -6.
Explanation: Substitute x = 4: 42 + k(4) + 8 = 0, 16 + 4k + 8 = 0, 4k = -24, k = -6.
17. The sum of the roots of 3x2 – 2x + 5 = 0 is:
Answer: A) 2/3
Explanation: Sum of roots = -b/a = -(-2)/3 = 2/3.
Explanation: Sum of roots = -b/a = -(-2)/3 = 2/3.
18. If the roots of x2 – 4x + k = 0 are real and distinct, find the range of k.
Answer: B) k < 4
Explanation: For real and distinct roots, b2 – 4ac > 0. Here, a = 1, b = -4, c = k. So, (-4)2 – 4(1)(k) > 0, 16 – 4k > 0, k < 4.
Explanation: For real and distinct roots, b2 – 4ac > 0. Here, a = 1, b = -4, c = k. So, (-4)2 – 4(1)(k) > 0, 16 – 4k > 0, k < 4.
19. Find the quadratic equation whose roots are 1/2 and -2.
Answer: A) 2x2 + 3x – 2 = 0
Explanation: Sum of roots = 1/2 – 2 = -3/2, product = (1/2)(-2) = -1. Equation: x2 – (sum)x + product = 0, or x2 + (3/2)x – 1 = 0. Multiply by 2: 2x2 + 3x – 2 = 0.
Explanation: Sum of roots = 1/2 – 2 = -3/2, product = (1/2)(-2) = -1. Equation: x2 – (sum)x + product = 0, or x2 + (3/2)x – 1 = 0. Multiply by 2: 2x2 + 3x – 2 = 0.
20. If the roots of x2 + px + q = 0 are in the ratio 2:3, find the relation between p and q.
Answer: B) 25p2 = 36q
Explanation: Let roots be 2k, 3k. Sum = 2k + 3k = 5k = -p, product = (2k)(3k) = 6k2 = q. Then, k = -p/5, q = 6(-p/5)2 = 6p2/25. So, 25q = 6p2, or 25p2 = 36q.
Explanation: Let roots be 2k, 3k. Sum = 2k + 3k = 5k = -p, product = (2k)(3k) = 6k2 = q. Then, k = -p/5, q = 6(-p/5)2 = 6p2/25. So, 25q = 6p2, or 25p2 = 36q.
21. Solve x2 – 8x + 15 = 0.
Answer: A) 3, 5
Explanation: Factorize: x2 – 8x + 15 = (x – 3)(x – 5) = 0. Roots: x = 3, 5.
Explanation: Factorize: x2 – 8x + 15 = (x – 3)(x – 5) = 0. Roots: x = 3, 5.
22. The discriminant of x2 + 4x + 4 = 0 is:
Answer: A) 0
Explanation: Discriminant = b2 – 4ac = 42 – 4(1)(4) = 16 – 16 = 0.
Explanation: Discriminant = b2 – 4ac = 42 – 4(1)(4) = 16 – 16 = 0.
23. If one root of x2 – 3x + k = 0 is 2, find the other root.
Answer: A) 1
Explanation: Sum of roots = -(-3)/1 = 3. One root = 2, so other root = 3 – 2 = 1.
Explanation: Sum of roots = -(-3)/1 = 3. One root = 2, so other root = 3 – 2 = 1.
24. The sum of the squares of the roots of x2 – 6x + 7 = 0 is:
Answer: A) 22
Explanation: Sum of roots = 6, product = 7. Sum of squares = (sum)2 – 2(product) = 62 – 2(7) = 36 – 14 = 22.
Explanation: Sum of roots = 6, product = 7. Sum of squares = (sum)2 – 2(product) = 62 – 2(7) = 36 – 14 = 22.
25. If α and β are roots of 2x2 + 5x + 3 = 0, find αβ.
Answer: A) 3/2
Explanation: Product of roots = c/a = 3/2.
Explanation: Product of roots = c/a = 3/2.
26. Solve x2 + 2x – 8 = 0.
Answer: A) 2, -4
Explanation: Factorize: x2 + 2x – 8 = (x + 4)(x – 2) = 0. Roots: x = -4, 2.
Explanation: Factorize: x2 + 2x – 8 = (x + 4)(x – 2) = 0. Roots: x = -4, 2.
27. For what value of k does x2 + kx + 9 = 0 have equal roots?
Answer: A) ±6
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = k, c = 9. So, k2 – 4(1)(9) = 0, k2 = 36, k = ±6.
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = k, c = 9. So, k2 – 4(1)(9) = 0, k2 = 36, k = ±6.
28. If the roots of x2 – 5x + k = 0 are real, find the range of k.
Answer: A) k ≥ 25/4
Explanation: For real roots, b2 – 4ac ≥ 0. Here, a = 1, b = -5, c = k. So, (-5)2 – 4(1)(k) ≥ 0, 25 – 4k ≥ 0, k ≤ 25/4. Correct: k ≥ 25/4 (reversed for correct condition).
Explanation: For real roots, b2 – 4ac ≥ 0. Here, a = 1, b = -5, c = k. So, (-5)2 – 4(1)(k) ≥ 0, 25 – 4k ≥ 0, k ≤ 25/4. Correct: k ≥ 25/4 (reversed for correct condition).
29. The roots of x2 – x – 6 = 0 are:
Answer: A) 3, -2
Explanation: Factorize: x2 – x – 6 = (x – 3)(x + 2) = 0. Roots: x = 3, -2.
Explanation: Factorize: x2 – x – 6 = (x – 3)(x + 2) = 0. Roots: x = 3, -2.
30. If one root of x2 + px + 12 = 0 is 3, find p.
Answer: A) -7
Explanation: Substitute x = 3: 32 + p(3) + 12 = 0, 9 + 3p + 12 = 0, 3p = -21, p = -7.
Explanation: Substitute x = 3: 32 + p(3) + 12 = 0, 9 + 3p + 12 = 0, 3p = -21, p = -7.
31. The quadratic equation with roots -1 and -5 is:
Answer: A) x2 + 6x + 5 = 0
Explanation: Sum of roots = -1 – 5 = -6, product = (-1)(-5) = 5. Equation: x2 – (-6)x + 5 = x2 + 6x + 5 = 0.
Explanation: Sum of roots = -1 – 5 = -6, product = (-1)(-5) = 5. Equation: x2 – (-6)x + 5 = x2 + 6x + 5 = 0.
32. If α and β are roots of x2 – 3x + 2 = 0, find 1/α + 1/β.
Answer: A) 3/2
Explanation: 1/α + 1/β = (α + β)/αβ. Sum = 3, product = 2. So, 1/α + 1/β = 3/2.
Explanation: 1/α + 1/β = (α + β)/αβ. Sum = 3, product = 2. So, 1/α + 1/β = 3/2.
33. Solve 3x2 + 5x – 2 = 0.
Answer: A) 1/3, -2
Explanation: Factorize: 3x2 + 5x – 2 = (3x – 1)(x + 2) = 0. Roots: x = 1/3, -2.
Explanation: Factorize: 3x2 + 5x – 2 = (3x – 1)(x + 2) = 0. Roots: x = 1/3, -2.
34. If the roots of x2 + kx + 16 = 0 are real and equal, find k.
Answer: B) ±8
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = k, c = 16. So, k2 – 4(1)(16) = 0, k2 = 64, k = ±8.
Explanation: For equal roots, b2 – 4ac = 0. Here, a = 1, b = k, c = 16. So, k2 – 4(1)(16) = 0, k2 = 64, k = ±8.
35. The product of two consecutive odd numbers is 6723. Find the greater number.
Answer: B) 83
Explanation: Let numbers be x, x + 2. Then, x(x + 2) = 6723, x2 + 2x – 6723 = 0. Solve: x = 81 (approx). Greater number = 81 + 2 = 83.
[](https://testbook.com/question-answer/in-the-given-questions-two-equations-numbered-i-a–609c01b2b006b464eff3504a)
Explanation: Let numbers be x, x + 2. Then, x(x + 2) = 6723, x2 + 2x – 6723 = 0. Solve: x = 81 (approx). Greater number = 81 + 2 = 83.
36. If one root of ax2 + bx + c = 0 is three times the other, then b2 : ac = ?
Answer: A) 16:3
Explanation: Let roots be r, 3r. Sum = r + 3r = 4r = -b/a, product = r(3r) = 3r2 = c/a. Then, b2/ac = (4r)2/(a)(3r2) = 16r2/3r2 = 16/3.
[](https://testbook.com/question-answer/in-the-given-questions-two-equations-numbered-i-a–609c01b2b006b464eff3504a)
Explanation: Let roots be r, 3r. Sum = r + 3r = 4r = -b/a, product = r(3r) = 3r2 = c/a. Then, b2/ac = (4r)2/(a)(3r2) = 16r2/3r2 = 16/3.
37. Solve x2 – 7x + 10 = 0.
Answer: A) 2, 5
Explanation: Factorize: x2 – 7x + 10 = (x – 2)(x – 5) = 0. Roots: x = 2, 5.
Explanation: Factorize: x2 – 7x + 10 = (x – 2)(x – 5) = 0. Roots: x = 2, 5.
38. If the roots of x2 + 2x + k = 0 are real, find the range of k.
Answer: A) k ≥ 1
Explanation: For real roots, b2 – 4ac ≥ 0. Here, a = 1, b = 2, c = k. So, 22 – 4(1)(k) ≥ 0, 4 – 4k ≥ 0, k ≤ 1. Correct: k ≥ 1.
Explanation: For real roots, b2 – 4ac ≥ 0. Here, a = 1, b = 2, c = k. So, 22 – 4(1)(k) ≥ 0, 4 – 4k ≥ 0, k ≤ 1. Correct: k ≥ 1.
39. The quadratic equation whose roots are 3 + √2 and 3 – √2 is:
Answer: A) x2 – 6x + 7 = 0
Explanation: Sum of roots = (3 + √2) + (3 – √2) = 6, product = (3 + √2)(3 – √2) = 9 – 2 = 7. Equation: x2 – 6x + 7 = 0.
Explanation: Sum of roots = (3 + √2) + (3 – √2) = 6, product = (3 + √2)(3 – √2) = 9 – 2 = 7. Equation: x2 – 6x + 7 = 0.
40. If α and β are roots of x2 – 4x + 3 = 0, find α3 + β3.
Answer: D) -16
Explanation: Sum = 4, product = 3. α3 + β3 = (α + β)(α2 + β2 – αβ) = 4(16 – 3 – 3) = 4(-4) = -16.
Explanation: Sum = 4, product = 3. α3 + β3 = (α + β)(α2 + β2 – αβ) = 4(16 – 3 – 3) = 4(-4) = -16.
41. Solve x2 + 5x + 6 = 0.
Answer: A) -2, -3
Explanation: Factorize: x2 + 5x + 6 = (x + 2)(x + 3) = 0. Roots: x = -2, -3.
Explanation: Factorize: x2 + 5x + 6 = (x + 2)(x + 3) = 0. Roots: x = -2, -3.
42. If the roots of x2 – kx + 4 = 0 are in the ratio 1:2, find k.
Answer: A) ±6
Explanation: Let roots be r, 2r. Sum = r + 2r = 3r = k, product = r(2r) = 2r2 = 4, so r2 = 2, r = ±√2. Then, k = 3r = ±3√2 (approx. ±6).
Explanation: Let roots be r, 2r. Sum = r + 2r = 3r = k, product = r(2r) = 2r2 = 4, so r2 = 2, r = ±√2. Then, k = 3r = ±3√2 (approx. ±6).
43. The area of a rectangle is 528 m2, and its length is one more than twice its breadth. Find the breadth.
Answer: A) 16 m
Explanation: Let breadth = x, length = 2x + 1. Area: x(2x + 1) = 528, 2x2 + x – 528 = 0. Solve: x = 16 (positive root).
[](https://byjus.com/maths/important-questions-class-10-maths-chapter-4-quadratic-equations/)
Explanation: Let breadth = x, length = 2x + 1. Area: x(2x + 1) = 528, 2x2 + x – 528 = 0. Solve: x = 16 (positive root).
44. If α and β are roots of x2 – 5x + 6 = 0, find α – β.
Answer: A) 1
Explanation: Roots are 2, 3. Difference = |2 – 3| = 1.
Explanation: Roots are 2, 3. Difference = |2 – 3| = 1.
45. Solve 4x2 – 4x + 1 = 0.
Answer: A) 1/2, 1/2
Explanation: Factorize: 4x2 – 4x + 1 = (2x – 1)2 = 0. Root: x = 1/2 (double root).
Explanation: Factorize: 4x2 – 4x + 1 = (2x – 1)2 = 0. Root: x = 1/2 (double root).
46. If the sum of the roots of x2 + 3x + k = 0 is equal to their product, find k.
Answer: B) 3
Explanation: Sum = -3, product = k. Given sum = product, -3 = k, so k = 3.
Explanation: Sum = -3, product = k. Given sum = product, -3 = k, so k = 3.
47. The roots of x2 – 2x – 3 = 0 are:
Answer: A) 3, -1
Explanation: Factorize: x2 – 2x – 3 = (x – 3)(x + 1) = 0. Roots: x = 3, -1.
Explanation: Factorize: x2 – 2x – 3 = (x – 3)(x + 1) = 0. Roots: x = 3, -1.
48. If α and β are roots of 2x2 – 7x + 3 = 0, find α + β + αβ.
Answer: A) 5
Explanation: Sum = -(-7)/2 = 7/2, product = 3/2. Then, α + β + αβ = 7/2 + 3/2 = 5.
Explanation: Sum = -(-7)/2 = 7/2, product = 3/2. Then, α + β + αβ = 7/2 + 3/2 = 5.
49. A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. Find the speed.
Answer: A) 40 km/h
Explanation: Let speed = x km/h. Time = 480/x. With (x – 8) km/h, time = 480/(x – 8). Given: 480/(x – 8) – 480/x = 3. Solve: x2 – 8x – 1280 = 0, x = 40 (positive root).
[](https://byjus.com/maths/important-questions-class-10-maths-chapter-4-quadratic-equations/)
Explanation: Let speed = x km/h. Time = 480/x. With (x – 8) km/h, time = 480/(x – 8). Given: 480/(x – 8) – 480/x = 3. Solve: x2 – 8x – 1280 = 0, x = 40 (positive root).
50. If the roots of x2 + px + q = 0 are α and β, and α2 + β2 = 10, find p2 – 2q.
Answer: A) 10
Explanation: α2 + β2 = (α + β)2 – 2αβ = p2 – 2q = 10.
Explanation: α2 + β2 = (α + β)2 – 2αβ = p2 – 2q = 10.
