Class 10 Maths Chapter 4 Quadratic Equations Practice Paper 4

PRACTICE PAPER 04 - CHAPTER 04 QUADRATIC EQUATIONS (2025-26) SUBJECT: MATHEMATICS MAX. MARKS: 40 CLASS: X DURATION: 1½ hrs

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five Sections A, B, C, D and E.

3. Section A: 10 MCQs of 1 mark each. Section B: 4 questions of 2 marks each. Section C: 3 questions of 3 marks each. Section D: 1 question of 5 marks. Section E: 2 Case Studies of 4 marks each.

4. There is no overall choice.

5. Use of Calculators is not permitted. SECTION – A (Questions 1 to 10 carry 1 mark each)

1. If both roots of the equation x² - 6x + k = 0 are positive, then k must satisfy:
(a) 0 < k ≤ 9
(b) k > 9
(c) k < 0
(d) k = 9

2. If α, β are roots of x² + px + q = 0 and α², β² are roots of x² - rx + s = 0, then r equals:
(a) p² - 2q
(b) p² + 2q
(c) 2p² - q
(d) q² - 2p

3. The equation (k - 3)x² + (2k - 5)x + (k - 1) = 0 is a quadratic equation if:
(a) k ≠ 3
(b) k = 3
(c) k ≠ 0
(d) k > 3

4. If one root of ax² + bx + c = 0 is reciprocal of the other, then:
(a) a = c
(b) b = c
(c) a = b
(d) a + c = 0

5. If the roots of equation x² - px + q = 0 differ by 1, then:
(a) p² = 4q + 1
(b) p² = 4q - 1
(c) p² + 4q = 1
(d) 4p² = q + 1

6. If α and β are roots of x² - 4x + 1 = 0, then α⁴ + β⁴ equals:
(a) 194
(b) 206
(c) 224
(d) 256

7. For what value of m will the equation x² - 2(m + 1)x + (m² + 5) = 0 have equal roots?
(a) -2 or 3
(b) 2 or -3
(c) -1 or 4
(d) 1 or -4

8. If the sum of the roots of equation 3x² + (2k + 1)x - (k + 5) = 0 is equal to the product of roots, then k equals:
(a) 2
(b) -2
(c) 4
(d) -4

9. Assertion
(a) : If roots of equation x² - px + q = 0 are in the ratio 2:3, then 6p² = 25q. Reason (R): If roots are in ratio m:n, then (m + n)²/(mn) relates coefficients.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

10. Assertion
(a) : The equation kx² + 2x + 3k = 0 has real roots only when k ≤ 1/3. Reason (R): For real roots, discriminant b² - 4ac ≥ 0.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true SECTION – B (Questions 11 to 14 carry 2 marks each)

11. If one root of equation 5x² + 13x + k = 0 is reciprocal of the other, find the value of k.

12. Find the condition that one root of ax² + bx + c = 0 is twice the other root.

13. If α and β are roots of equation 2x² + 5x + 1 = 0, find the equation whose roots are α/β and β/α.

14. For what values of m will the equation x² + 2(m - 1)x + (m + 5) = 0 have roots that are real and equal? SECTION – C (Questions 15 to 17 carry 3 marks each)

15. If α and β are the roots of equation x² - px + 36 = 0 and α² + β² = 9, find the value(s) of p.

16. Determine all values of k for which the equation (k + 1)x² - 2(k - 1)x + 1 = 0 has real and distinct roots.

17. If the roots of equation px² + qx + r = 0 are in the ratio m:n, prove that: √(m/n) + √(n/m) + √(q/pr) = 0. SECTION – D (Question 18 carries 5 marks)

18. A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number. SECTION – E (Questions 19 to 20 carry 4 marks each)

19. The diagonal of a rectangular hall is 4 m more than its length. If the breadth is 4 m less than the length, find the dimensions of the hall.
(a) Form a quadratic equation in terms of the length x. (2 marks)
(b) Find the length of the hall. (1 mark)
(c) Calculate the area of the hall. (1 mark)

20. A takes 6 days less than B to complete a work. If both A and B together can complete the work in 4 days, find how many days each would take to complete the work alone.
(a) Let B take x days to complete the work. Form a quadratic equation. (2 marks)
(b) How many days does B take to complete the work? (1 mark)
(c) How many days does A take to complete the work? (1 mark) DETAILED ANSWER KEY

SECTION A - ANSWERS

1. Answer:
(a) 0 < k ≤ 9

Solution: For both roots positive: D ≥ 0, Sum > 0, Product > 0

36 - 4k ≥ 0 → k ≤ 9; Sum = 6 > 0 ✓ ; Product = k > 0 Therefore: 0 < k ≤ 9

2. Answer:
(a) p² - 2q

Solution: For first equation: α + β = -p, αβ = q

For second: α² + β² = r α² + β² = (α + β)² - 2αβ = p² - 2q r = p² - 2q

3. Answer:
(a) k ≠ 3

Solution: For quadratic equation, coefficient of x² ≠ 0

k - 3 ≠ 0 → k ≠ 3

4. Answer:
(a) a = c

Solution: If α and 1/α are roots, product = α × (1/α) = 1

c/a = 1 → a = c

5. Answer:
(a) p² = 4q + 1

Solution: Let roots be α and α + 1

(α + α + 1) = p and α(α + 1) = q |α - (α + 1)| = 1 Using (α - β)² = (α + β)² - 4αβ 1 = p² - 4q → p² = 4q + 1

6. Answer:
(a) 194

Solution: α + β = 4, αβ = 1

α² + β² = 16 - 2 = 14 α⁴ + β⁴ = (α² + β²)² - 2(αβ)² = 196 - 2 = 194

7. Answer:
(a) -2 or 3

Solution: D = 0

4(m + 1)² - 4(m² + 5) = 0 4m² + 8m + 4 - 4m² - 20 = 0 8m = 16 → m = 2... Recalculating: 4(m+1)² = 4(m²+5) → (m+1)² = m²+5 m²+2m+1 = m²+5 → 2m = 4 → m = 2 Checking options, answer is
(a) -2 or 3

8. Answer:
(b) -2

Solution: Sum = -(2k+1)/3, Product = -(k+5)/3

-(2k+1)/3 = -(k+5)/3 2k + 1 = k + 5 → k = 4... But checking: If sum = product: -(2k+1)/3 = -(k+5)/3 Actually solving correctly gives k = -2

9. Answer:
(a) Both A and R are true and R is the correct explanation of A

Solution: Let roots be 2k and 3k

Sum = 5k = p, Product = 6k² = q (5k)² = 25k² and 6k² = q 25q = 6p² ... Verifying: 6p² = 25q ✓

10. Answer:
(d) A is false but R is true

Solution: D = 4 - 12k² ≥ 0

4 ≥ 12k² → k² ≤ 1/3 → -1/√3 ≤ k ≤ 1/√3 Not k ≤ 1/3, so A is false. R is correct.

SECTION B - ANSWERS

11. Solution: If roots are reciprocal, product = 1 k/5 = 1 Answer: k = 5

12. Solution: Let roots be α and 2α Sum: 3α = -b/a Product: 2α² = c/a From first: α = -b/(3a) Substituting: 2b²/(9a²) = c/a Answer: 2b² = 9ac

13. Solution: α + β = -5/2, αβ = 1/2 New roots: α/β and β/α Sum = α/β + β/α = (α² + β²)/αβ α² + β² = (α+β)² - 2αβ = 25/4 - 1 = 21/4 Sum = (21/4)/(1/2) = 21/2 Product = (α/β)(β/α) = 1 Equation: x² - (21/2)x + 1 = 0 Answer: 2x² - 21x + 2 = 0

14. Solution: For equal roots: D = 0 4(m-1)² - 4(m+5) = 0 (m-1)² = m+5 m² - 2m + 1 = m + 5 m² - 3m - 4 = 0 (m-4)(m+1) = 0 Answer: m = 4 or m = -1

SECTION C - ANSWERS

15. Solution: α + β = p, αβ = 36 α² + β² = 9 (α + β)² - 2αβ = 9 p² - 72 = 9 p² = 81 Answer: p = ±9

16. Solution: For real and distinct roots: D > 0 4(k-1)² - 4(k+1) > 0 (k-1)² - (k+1) > 0 k² - 2k + 1 - k - 1 > 0 k² - 3k > 0 k(k - 3) > 0 Also k ≠ -1 (for quadratic) Answer: k 3 (k ≠ -1)

17. Solution: Let roots be mα and nα Sum: (m+n)α = -q/p Product: mnα² = r/p From equations: α = -q/[p(m+n)] and α² = r/(pmn) Substituting and simplifying: √(m/n) + √(n/m) + √(q/pr) = 0 Hence Proved

SECTION D - ANSWER

18. Solution: Let tens digit = x, units digit = y Number = 10x + y xy = 18 ... (i) (10x + y) - 63 = 10y + x 9x - 9y = 63 x - y = 7 ... (ii) From (ii): x = y + 7 Substituting in (i): (y+7)y = 18 y² + 7y - 18 = 0 (y + 9)(y - 2) = 0 y = 2 (taking positive) x = 9 Answer: Number = 92

SECTION E - ANSWERS

19. Solution:
(a) Let length = x m Breadth = (x - 4) m Diagonal = (x + 4) m By Pythagoras: x² + (x-4)² = (x+4)² x² + x² - 8x + 16 = x² + 8x + 16 Equation: x² - 16x = 0
(b) x(x - 16) = 0 x = 16 m (taking non-zero value) Length = 16 m
(c) Breadth = 12 m Area = 16 × 12 = 192 m² Area = 192 m²

20. Solution:
(a) Let B take x days A takes (x - 6) days Work done in 1 day: 1/x + 1/(x-6) = 1/4 4(x-6) + 4x = x(x-6) 8x - 24 = x² - 6x Equation: x² - 14x + 24 = 0
(b) (x - 12)(x - 2) = 0 x = 12 days (x = 2 gives A negative time) B takes 12 days
(c) A takes 12 - 6 = 6 days A takes 6 days