CBSE Class 10 Mathematics Quadratic Equations MCQs Set 16

Answer: (d) Suman

Explanation:
1. Since q² − 4pr > 0, the two roots are real and distinct.
2. If p, q, r are all positive, the sum of the roots is \( -\frac{q}{p} < 0 \) and the product is \( \frac{r}{p} > 0 \).
3. A negative sum with a positive product means both roots are negative, so Suman is correct.

Teacher's Note:
1. The sign of the sum and the product tells us the signs of both roots.
2. Zain and Vipul ignore the sign of \( -\frac{q}{p} \).

Answer: (b) 2

Explanation:
1. Put \( x = \frac{1}{2} \): \( \frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0 \).
2. So \( \frac{k}{2} = 1 \), hence k = 2.

Teacher's Note:
1. A root satisfies the equation, so substitute it directly.
2. Clear the fractions at the end to avoid sign slips.

Answer: (a) x² + 10x − 500 = 0

Explanation:
1. Share per person = \( \frac{4000}{x} \); with 10 more persons it is \( \frac{4000}{x+10} \).
2. \( \frac{4000}{x} - \frac{4000}{x+10} = 80 \Rightarrow \frac{40000}{x(x+10)} = 80 \).
3. So x(x + 10) = 500, i.e. x² + 10x − 500 = 0.

Teacher's Note:
1. Write the share before and after the change, then subtract.
2. Always simplify to the standard form ax² + bx + c = 0.

Answer: (c) 4

Explanation:
1. Let \( y = \sqrt{12+\sqrt{12+\cdots}} \). Then \( y = \sqrt{12+y} \), so y² = 12 + y.
2. y² − y − 12 = 0 gives (y − 4)(y + 3) = 0.
3. Since y > 0, y = 4.

Teacher's Note:
1. Repeating radicals are solved by letting the whole expression equal y.
2. Reject the negative root because a square root is not negative.

Answer: (b) 7 years

Explanation:
1. Let the present age be x. Then \( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \).
2. \( 3(2x+2) = (x-3)(x+5) \Rightarrow x^2 - 4x - 21 = 0 \).
3. (x − 7)(x + 3) = 0, so x = 7 years.

Teacher's Note:
1. Age 3 years ago is x − 3 and age 5 years later is x + 5.
2. Ages cannot be negative, so reject x = −3.

Answer: (c) 18, 12

Explanation:
1. Let the larger number be x. The square of the smaller is 8x.
2. x² − 8x = 180, so x² − 8x − 180 = 0, i.e. (x − 18)(x + 10) = 0.
3. x = 18 (the larger number is positive). The smaller number squared is 144, so the smaller number is 12.

Teacher's Note:
1. Translate each sentence into an equation using one variable.
2. Check: 18² − 12² = 324 − 144 = 180.

Answer: (d) 2x² − 7x + 6 = 0

Explanation:
1. Substitute x = 2 in each option.
2. (a) 4 − 8 + 5 = 1, (b) 4 + 6 − 12 = −2, (d) 12 − 12 − 2 = −2, all not zero.
3. (c) 8 − 14 + 6 = 0, so 2 is a root of option (c).

Teacher's Note:
1. A number is a root only if the equation becomes 0.
2. Test every option before choosing.

Answer: (a) x² + x − 5 = 0

Explanation:
1. (a) D = 18 − 18 = 0 (equal roots).
2. (b) D = 1 + 20 = 21 > 0 (distinct real roots).
3. (c) D = 9 − 8√2 < 0 since 8√2 ≈ 11.3. (d) D = 9 − 20 < 0.

Teacher's Note:
1. Compute D = b² − 4ac for each option.
2. D > 0 gives two distinct real roots.

Answer: (b) no real roots

Explanation:
1. D = (−√5)² − 4(2)(1) = 5 − 8 = −3.
2. D < 0, so the equation has no real roots.

Teacher's Note:
1. The sign of D decides the nature of the roots.
2. A quadratic never has more than two roots.

Answer: (a) 0, 8

Explanation:
1. Equal roots need D = 0: k² − 4(2)(k) = 0.
2. k² − 8k = 0, so k(k − 8) = 0.
3. Hence k = 0 or k = 8.

Teacher's Note:
1. Do not divide by k; factorise instead, otherwise k = 0 is lost.
2. Both values give equal roots.

Answer: (d) k ≤ 16

Explanation:
1. Real roots need D ≥ 0: 64 − 4k ≥ 0.
2. So k ≤ 16.

Teacher's Note:
1. Real roots include the equal-roots case, so use ≥.
2. Isolate k after writing D.

Answer: (c) 2, −5

Explanation:
1. Factorise: x² + 5x − 2x − 10 = x(x + 5) − 2(x + 5) = (x + 5)(x − 2).
2. So x = −5 or x = 2.

Teacher's Note:
1. Find two numbers with product −10 and sum 3.
2. Check: sum of roots = −3 and product = −10.

Answer: (d) x² − 16

Explanation:
1. Sum of zeroes is 0, so the other zero is −4.
2. Polynomial: x² − (sum)x + product = x² − 0·x + (4)(−4) = x² − 16.

Teacher's Note:
1. Use x² − (sum of zeroes)x + product of zeroes.
2. Zeroes of x² − 16 are ±4.

Answer: (c) 0 and 2

Explanation:
1. 3x² − 6x = 0, so 3x(x − 2) = 0.
2. x = 0 or x = 2.

Teacher's Note:
1. Never cancel x on both sides; you would lose the root x = 0.
2. Take everything to one side and factorise.

Answer: (a) −m, m + 3

Explanation:
1. x² − 3x − m(m + 3) = x² + mx − (m + 3)x − m(m + 3).
2. = x(x + m) − (m + 3)(x + m) = (x + m)(x − m − 3).
3. So x = −m or x = m + 3.

Teacher's Note:
1. Split the middle term −3x as mx − (m + 3)x.
2. Check: sum = 3 and product = −m(m + 3).

Answer: (b) \( -\frac{2}{a} \)

Explanation:
1. Since α + β = −b/a, we get b = −a(α + β).
2. So aβ + b = aβ − aα − aβ = −aα, and aα + b = −aβ.
3. The expression becomes \( \frac{\alpha}{-a\alpha} + \frac{\beta}{-a\beta} = -\frac{1}{a} - \frac{1}{a} = -\frac{2}{a} \).

Teacher's Note:
1. Replace b using the sum of the roots.
2. The α and β cancel neatly in each fraction.

Answer: (c) x² + 51x − 156 = 0

Explanation:
1. Let Aditya's present age be x; his father's age is x + 45.
2. After 3 years the ages are x + 3 and x + 48: (x + 3)(x + 48) = 300.
3. x² + 51x + 144 − 300 = 0, i.e. x² + 51x − 156 = 0.

Teacher's Note:
1. Add 3 to both ages before multiplying.
2. Expand carefully: 3 × 48 = 144.

Answer: (a) x = 24, x = −6

Explanation:
1. \( \frac{x^2}{9}+\frac{x^2}{2} = \frac{11x^2}{18} \), so \( \frac{11x^2}{18} + 7x + 56 = x^2 \).
2. Multiply by 18: 11x² + 126x + 1008 = 18x², i.e. 7x² − 126x − 1008 = 0, so x² − 18x − 144 = 0.
3. (x − 24)(x + 6) = 0, so x = 24 or x = −6.

Teacher's Note:
1. Combine the x² terms first, then clear fractions.
2. Check the sum of roots: 24 + (−6) = 18.

Answer: (b) −9 < λ < 7

Explanation:
1. No real roots need D < 0: (λ + 1)² − 4(8)(2) < 0.
2. (λ + 1)² < 64, so −8 < λ + 1 < 8.
3. Hence −9 < λ < 7.

Teacher's Note:
1. Do not forget the negative square root when solving a square inequality.
2. Subtract 1 from all three parts at the end.

Answer: (d) No, because 0.2² − 0.4 = −0.36 ≠ 0

Explanation:
1. Put x = 0.2: (0.2)² − 0.4 = 0.04 − 0.4 = −0.36.
2. The value is not 0, so 0.2 is not a root.

Teacher's Note:
1. A root must make the left side exactly 0.
2. Do not confuse 0.2² = 0.04 with 0.4.

Answer: (c) 12 cm

Explanation:
1. Let the base be x cm; the altitude is (x − 7) cm.
2. x² + (x − 7)² = 169 gives 2x² − 14x − 120 = 0, i.e. x² − 7x − 60 = 0.
3. (x − 12)(x + 5) = 0, so x = 12. The sides are 12 cm and 5 cm.

Teacher's Note:
1. Use the Pythagoras theorem to form the equation.
2. Check: 12² + 5² = 144 + 25 = 169.

Answer: (d) real and equal, each \( \frac{2}{\sqrt{3}} \)

Explanation:
1. D = (4√3)² − 4(3)(4) = 48 − 48 = 0, so the roots are real and equal.
2. Each root is \( \frac{-b}{2a} = \frac{4\sqrt{3}}{6} = \frac{2}{\sqrt{3}} \).

Teacher's Note:
1. When D = 0, each root equals −b/(2a).
2. The equation is also (√3x − 2)² = 0.

Answer: (a) \( -\frac{1}{4} \)

Explanation:
1. Put x = 2: 4k + 4 − 3 = 0, so 4k = −1.
2. Hence \( k = -\frac{1}{4} \).

Teacher's Note:
1. Substitute the given solution and solve the linear equation in k.
2. Check the sign of the constant terms: 4 − 3 = 1.

Answer: (b) \( a > \frac{3}{10} \)

Explanation:
1. No real roots need D < 0: 36 − 4(30a)(1) < 0.
2. 36 − 120a < 0, so \( a > \frac{36}{120} = \frac{3}{10} \).

Teacher's Note:
1. D < 0 means no real roots; D = 0 would give equal roots.
2. Divide by the positive number 120, so the inequality keeps its direction.

Answer: (b) \( \frac{9}{2} \) and 2

Explanation:
1. For ax² + bx + c = 0, sum = −b/a and product = c/a.
2. Sum = 9/2 and product = 4/2 = 2.
3. Check: roots are 4 and 1/2, with sum 9/2 and product 2.

Teacher's Note:
1. Take care with the sign: the sum is −b/a.
2. Verify by factorising: (2x − 1)(x − 4) = 0.

Answer: (d) 80, two distinct real roots

Explanation:
1. Here a = 4, b = 0, c = −5.
2. D = 0 − 4(4)(−5) = 80 > 0.
3. So the roots are real and distinct: \( x = \pm\frac{\sqrt{5}}{2} \).

Teacher's Note:
1. A missing x term means b = 0.
2. Positive D gives two distinct real roots.

Answer: (c) 14 years

Explanation:
1. Let the present age be x. Then (x − 5)² = 5x + 11.
2. x² − 10x + 25 = 5x + 11, so x² − 15x + 14 = 0.
3. (x − 14)(x − 1) = 0. x = 1 is rejected as x − 5 would be negative, so x = 14.

Teacher's Note:
1. Reject roots that give an impossible age.
2. Check: 9² = 81 and 5 × 14 + 11 = 81.

Answer: (a) p = −8, other root 2

Explanation:
1. Put x = 2/3: \( \frac{4}{3} + \frac{2p}{3} + 4 = 0 \), so \( \frac{2p}{3} = -\frac{16}{3} \) and p = −8.
2. Product of roots = 4/3, so the other root is \( \frac{4/3}{2/3} = 2 \).

Teacher's Note:
1. Use the product of roots c/a to get the other root quickly.
2. Check the sum: 2/3 + 2 = 8/3 = −p/3.

Answer: (d) 3

Explanation:
1. α + β = 5 and α − β = 1, so α = 3 and β = 2.
2. Product αβ = 6 = 3(k − 1), so k − 1 = 2.
3. Hence k = 3.

Teacher's Note:
1. Use the sum and the difference to find both roots first.
2. Then use the product of roots to find k.

Answer: (a) \( (3 - \sqrt{5}) \) m

Explanation:
1. Let AC = x, so CB = 2 − x. Then x² = 2(2 − x), i.e. x² + 2x − 4 = 0.
2. x = −1 ± √5; since x > 0, AC = (√5 − 1) m.
3. CB = 2 − (√5 − 1) = (3 − √5) m.

Teacher's Note:
1. Choose the positive root because a length cannot be negative.
2. Check: AC² ≈ 1.528 and 2 × CB ≈ 1.528.

Answer: (c) 3

Explanation:
1. Let the roots be r and 6r. Sum: 7r = 14/p, so r = 2/p.
2. Product: 6r² = 8/p, so 6 × 4/p² = 8/p, i.e. 24/p² = 8/p.
3. So p = 3.

Teacher's Note:
1. Use the sum and the product of roots together.
2. Check with p = 3: 3x² − 14x + 8 = (3x − 2)(x − 4), with roots 2/3 and 4, and 4 is 6 times 2/3.

Answer: (b) 40 km/h

Explanation:
1. Let the speed be x km/h: \( \frac{480}{x-8} - \frac{480}{x} = 3 \).
2. \( 480 \times 8 = 3x(x-8) \), so x² − 8x − 1280 = 0.
3. (x − 40)(x + 32) = 0, so x = 40 km/h.

Teacher's Note:
1. Subtract the two times and equate to the extra time.
2. Check: 480/32 − 480/40 = 15 − 12 = 3.

Answer: (b) 6 km/h

Explanation:
1. Let the stream speed be x km/h. \( \frac{24}{18-x} - \frac{24}{18+x} = 1 \).
2. 24(2x) = 324 − x², so x² + 48x − 324 = 0.
3. (x − 6)(x + 54) = 0, so x = 6 km/h.

Teacher's Note:
1. Upstream speed = 18 − x; downstream speed = 18 + x.
2. Reject the negative root.

Answer: (c) 14

Explanation:
1. Minutes left for 3 pm = 60 − t. So \( 60 - t = \frac{t^2}{4} - 3 \).
2. t² + 4t − 252 = 0, i.e. (t − 14)(t + 18) = 0.
3. t = 14.

Teacher's Note:
1. The time from 2:t to 3:00 is 60 − t minutes.
2. Reject the negative root.

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

Explanation:
1. For 4x² − 12x + 9 = 0, D = 144 − 4(4)(9) = 144 − 144 = 0.
2. So the roots are real and equal, and Assertion is true.
3. Reason is the standard rule, and it explains the Assertion.

Teacher's Note:
1. D = 0 always means a repeated real root.
2. Here (2x − 3)² = 0 confirms it.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Put \( x = \frac{2}{3} \): \( 6 \cdot \frac{4}{9} - \frac{2}{3} - k = 0 \Rightarrow \frac{8}{3} - \frac{2}{3} = k \), so k = 2. Assertion is true.
2. A quadratic equation always has two roots, so Reason is also true.
3. But this fact does not help us find k, so Reason is not the explanation of Assertion.

Teacher's Note:
1. Both statements can be true without one explaining the other.
2. Ask: does the Reason actually give the value of k? Here it does not.

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

Explanation:
1. For x² + 2x + 2 = 0, D = 4 − 8 = −4 < 0.
2. So the equation has no real roots; Assertion is true.
3. Reason is the rule that gives this conclusion, so it is the correct explanation.

Teacher's Note:
1. A negative discriminant means no real roots.
2. Check D before trying to solve.

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

Explanation:
1. Equal roots need D = 0: (3k)² − 4(8)(2) = 0, so 9k² = 64.
2. k = ±8/3, so Assertion is true.
3. Reason is the condition used, so it explains Assertion.

Teacher's Note:
1. Set D = 0 and solve for k.
2. Remember both signs of the square root.

Answer: (c) A is true, but R is false.

Explanation:
1. 8x² − 16x + 8 = 8(x − 1)², so the roots are 1, 1 (repeated). Assertion is true.
2. Repeated roots occur when D = 0, not D > 0, so Reason is false.

Teacher's Note:
1. D > 0 gives two distinct roots; D = 0 gives repeated roots.
2. Always check each statement separately.

Answer: (c) A is true, but R is false.

Explanation:
1. Let \( t = \sqrt{\frac{\alpha}{1-\alpha}} \). Then \( t + \frac{1}{t} = \frac{13}{6} \), so 6t² − 13t + 6 = 0 and t = 3/2 or 2/3.
2. These give α = 9/13 or 4/13, both between 0 and 1, so real solutions exist and Assertion is true.
3. A quadratic equation can have two real roots, equal roots or no real roots, so Reason is false.

Teacher's Note:
1. A substitution turns the equation into a quadratic in t.
2. Reason is false because D < 0 gives no real root.