CBSE Class 10 Mathematics Quadratic Equations MCQs Set 17

Answer: (c) 8

Explanation:
1. Write the equation as \( x^2 - 3x + (k - 10) = 0 \).
2. Product of roots = \( k - 10 \) = −2.
3. So \( k = 8 \).

Teacher's Note:
1. The product of the roots of \( ax^2 + bx + c = 0 \) is \( \frac{c}{a} \).
2. Bring every term to the left side before reading off the constant term.

Answer: (d) 5

Explanation:
1. Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \).
2. Product of roots = 1, and it equals \( \frac{k}{5} \).
3. So \( k = 5 \).

Teacher's Note:
1. Reciprocal roots have product 1.
2. Equate \( \frac{c}{a} \) to 1.

Answer: (a) \( x^2 - 6x + 6 = 0 \)

Explanation:
1. The equation is \( x^2 - (\text{sum})x + \text{product} = 0 \).
2. So it is \( x^2 - 6x + 6 = 0 \).

Teacher's Note:
1. Remember the sign of the middle term is opposite to the sum.
2. Both the sum and the product here are positive.

Answer: (a) 1 or 4

Explanation:
1. Equal roots mean \( D = 0 \): \( 4(k + 2)^2 - 36k = 0 \).
2. So \( (k + 2)^2 = 9k \), i.e. \( k^2 - 5k + 4 = 0 \).
3. \( (k - 1)(k - 4) = 0 \), so \( k = 1 \) or \( k = 4 \).

Teacher's Note:
1. Set the discriminant equal to zero.
2. Divide by 4 early to keep the numbers small.

Answer: (d) either \( k > 2\sqrt{5} \) or \( k < -2\sqrt{5} \)

Explanation:
1. Real and distinct roots need \( D > 0 \): \( k^2 - 20 > 0 \).
2. So \( k^2 > 20 \), which means \( k > 2\sqrt{5} \) or \( k < -2\sqrt{5} \).

Teacher's Note:
1. A square greater than 20 allows both signs of k.
2. Do not forget the negative side.

Answer: (b) \( \frac{c}{b} \)

Explanation:
1. Product of roots: \( (a - b)(b - c) = \)\( \frac{c}{a} \).
2. Sum of roots: \( (a - b) + (b - c) = a - c = -\frac{b}{a} \), so \( c - a = \)\( \frac{b}{a} \).
3. The value is \( \frac{c/a}{b/a} \) = \( \frac{c}{b} \).

Teacher's Note:
1. Use both the sum and the product of the roots.
2. Note that c − a is the negative of the sum of the roots.

Answer: (d) −16

Explanation:
1. Product of roots = \( \frac{c}{a} \) = \( \frac{-16}{1} \) = −16.

Teacher's Note:
1. Read a, b and c with their signs.
2. No need to solve the equation.

Answer: (b) \( ax^2 - bx + c = 0 \)

Explanation:
1. New sum = \( -(\alpha + \beta) = \)\( \frac{b}{a} \), new product = \( \alpha\beta = \)\( \frac{c}{a} \).
2. Equation: \( x^2 - \frac{b}{a}x + \frac{c}{a} = 0 \), i.e. \( ax^2 - bx + c = 0 \).

Teacher's Note:
1. Changing the sign of both roots changes only the sign of the middle term.
2. Replace x by −x in the original equation to check.

Answer: (b) \( c^2 = a^2(1 + m^2) \)

Explanation:
1. Equal roots: \( (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0 \).
2. \( 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 = 0 \), so \( c^2 = a^2(1 + m^2) \).

Teacher's Note:
1. Expand carefully; the m²c² terms cancel.
2. The discriminant must be zero for equal roots.

Answer: (b) \( \frac{1}{b^2} \)

Explanation:
1. Group the terms: \( b^2x(a^2x + 1) - 1(a^2x + 1) = 0 \).
2. \( (a^2x + 1)(b^2x - 1) = 0 \), so \( x = -\frac{1}{a^2} \) or \( x = \frac{1}{b^2} \).
3. Only \( \frac{1}{b^2} \) appears in the options.

Teacher's Note:
1. Factorise by grouping the first and third, and second and fourth terms.
2. Check each option by substitution if unsure.

Answer: (a) ± 0.2

Explanation:
1. \( x^2 = 0.04 \), so \( x = \pm 0.2 \).

Teacher's Note:
1. Take the square root on both sides.
2. Remember both the positive and negative roots.

Answer: (d) 6, −1

Explanation:
1. The first student's sum of roots is correct: 3 + 2 = 5, so the coefficient of x is −5.
2. The second student's data: \( x^2 \) coefficient 1, constant −6.
3. Correct equation: \( x^2 - 5x - 6 = 0 \), i.e. \( (x - 6)(x + 1) = 0 \).
4. Roots are 6 and −1.

Teacher's Note:
1. The wrong constant does not affect the sum of the roots.
2. Combine the correct parts of the two students' work.

Answer: (b) \( 2b^2 = 9ac \)

Explanation:
1. Let the roots be \( \alpha \) and \( 2\alpha \). Then \( 3\alpha = -\frac{b}{a} \) and \( 2\alpha^2 = \frac{c}{a} \).
2. So \( 2\left(\frac{b}{3a}\right)^2 = \frac{c}{a} \), i.e. \( 2b^2 = 9ac \).

Teacher's Note:
1. Use the sum and the product of the roots together.
2. Eliminate α between the two relations.

Answer: (c) 4

Explanation:
1. Use \( A^2 - B^2 = (A - B)(A + B) \): \( (1)(2x - 2) = x + 2 \).
2. So \( 2x - 2 = x + 2 \) and \( x = 4 \).

Teacher's Note:
1. The x² terms cancel, so the equation is linear.
2. Use the difference of squares identity.

Answer: (c) {3, 4, −3, −4}

Explanation:
1. Put \( y = \frac{12}{x} \): \( x^2 + \frac{144}{x^2} = 25 \), so \( x^4 - 25x^2 + 144 = 0 \).
2. \( (x^2 - 16)(x^2 - 9) = 0 \), so \( x = \pm 4 \) or \( x = \pm 3 \).

Teacher's Note:
1. Substitute y in terms of x and treat x² as the unknown.
2. Each square gives two roots, positive and negative.

Answer: (a) 4

Explanation:
1. \( 7 + 4\sqrt{3} = (2 + \sqrt{3})^2 \), so \( x = 2 + \sqrt{3} \).
2. \( \frac{1}{x} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} \).
3. So \( x + \frac{1}{x} = 4 \).

Teacher's Note:
1. Write the number under the root as a perfect square.
2. Rationalise the denominator to find 1/x.

Answer: (b) \( q^2 = pr \)

Explanation:
1. First equation real: \( 4q^2 - 4pr \ge 0 \), so \( q^2 \ge pr \).
2. Second equation real: \( 4pr - 4q^2 \ge 0 \), so \( pr \ge q^2 \).
3. Both hold only if \( q^2 = pr \).

Teacher's Note:
1. Write the discriminant condition for each equation.
2. Two opposite inequalities force equality.

Answer: (d) Real

Explanation:
1. \( D = 4(p + 1)^2 - 8p = 4(p^2 + 1) \).
2. Since \( p^2 + 1 > 0 \) for every real p, D is always positive, so the roots are real (and distinct).
3. They need not be irrational: for \( p = 0 \) the roots are 0 and −1.

Teacher's Note:
1. Expand the discriminant and look for a sum of squares.
2. A positive discriminant guarantees real roots, not irrational ones.

Answer: (d) \( b^2r = q^2c \)

Explanation:
1. For roots in ratio \( k:1 \), \( \frac{b^2}{c} = \frac{(k + 1)^2}{k} \) is fixed by the ratio.
2. So \( \frac{b^2}{c} = \frac{q^2}{r} \), i.e. \( b^2r = q^2c \).

Teacher's Note:
1. The quantity (sum)² ÷ product depends only on the ratio of the roots.
2. Equate it for the two equations.

Answer: (d) 1, −8

Explanation:
1. Put \( y = x^{1/3} \): \( y^2 + y - 2 = 0 \), so \( (y - 1)(y + 2) = 0 \).
2. \( y = 1 \) or \( y = -2 \), so \( x = 1 \) or \( x = (-2)^3 = -8 \).

Teacher's Note:
1. Reduce to a quadratic by substitution.
2. Cube y to return to x.

Answer: (b) \( x(x + 1) + 1 = (x - 2)(x - 5) \)

Explanation:
1. Option (b): \( x^2 + x + 1 = x^2 - 7x + 10 \) gives \( 8x - 9 = 0 \), which is linear.
2. Options (a), (c) and (d) simplify to \( x^2 - 4x + 6 = 0 \), \( x^2 - 11x + 8 = 0 \) and \( 2x^2 - 13x + 9 = 0 \) (all of degree 2).

Teacher's Note:
1. Simplify each equation and check whether the x² term survives.
2. In (b) the x² terms cancel.

Answer: (d) \( a = c \)

Explanation:
1. Roots \( \alpha \) and \( \frac{1}{\alpha} \) have product 1 = \( \frac{c}{a} \).
2. So \( a = c \).

Teacher's Note:
1. Product of reciprocal roots is 1.
2. Use c/a = 1.

Answer: (b) \( 3, \frac{1}{3} \)

Explanation:
1. \( x + \frac{1}{x} = \frac{10}{3} \), so \( 3x^2 - 10x + 3 = 0 \).
2. \( (x - 3)(3x - 1) = 0 \), so \( x = 3 \) or \( x = \frac{1}{3} \).

Teacher's Note:
1. Convert the mixed fraction first.
2. Multiply by 3x to clear the denominators.

Answer: (b) \( mq = np \)

Explanation:
1. Equal roots: \( 4(mp + nq)^2 - 4(m^2 + n^2)(p^2 + q^2) = 0 \).
2. Expanding gives \( m^2q^2 + n^2p^2 - 2mnpq = 0 \), i.e. \( (mq - np)^2 = 0 \).
3. So \( mq = np \).

Teacher's Note:
1. Expand both products; the m²p² and n²q² terms cancel.
2. A perfect square equal to zero gives a single condition.

Answer: (a) \( b = 6,\ c = 9 \)

Explanation:
1. \( x^2 - 2x + 1 = (x - 1)^2 \), so the new roots are 1 and 1.
2. The original roots were 3 and 3, so \( b = 3 + 3 = 6 \) and \( c = 3 \times 3 = 9 \).

Teacher's Note:
1. Add 2 back to each new root to get the original roots.
2. Use sum and product to read b and c.

Answer: (c) \( (x + 2)(x - 1) = x^2 - 2x - 3 \)

Explanation:
1. Option (c): \( x^2 + x - 2 = x^2 - 2x - 3 \) gives \( 3x + 1 = 0 \), which is linear.
2. The others reduce to \( x^2 + 2x - 2 = 0 \), \( 10x^2 - 3x + 1 = 0 \) and \( 4x^2 + x = 0 \).

Teacher's Note:
1. Expand both sides and cancel like terms.
2. A quadratic equation must keep an x² term.

Answer: (b) \( \frac{3}{64} \)

Explanation:
1. Divide by 4: \( x^2 - \frac{\sqrt{3}}{4}x - \frac{5}{4} = 0 \).
2. Half of the coefficient of x is \( \frac{\sqrt{3}}{8} \), and its square is \( \frac{3}{64} \).
3. So \( \frac{3}{64} \) is added and subtracted.

Teacher's Note:
1. First make the coefficient of x² equal to 1.
2. Add and subtract the square of half the coefficient of x.

Answer: (c) \( 2x^2 - 7x + 6 = 0 \)

Explanation:
1. Put \( x = 2 \) in each equation.
2. Only (c) gives \( 2(4) - 7(2) + 6 = 8 - 14 + 6 = 0 \).

Teacher's Note:
1. A root makes the equation true.
2. Substitute the value in every option.

Answer: (d) 0, 8

Explanation:
1. \( D = k^2 - 8k = 0 \), so \( k(k - 8) = 0 \).
2. Hence \( k = 0 \) or \( k = 8 \).

Teacher's Note:
1. Do not divide by k; that would lose the root k = 0.
2. Factorise instead.

Answer: (c) \( x^2 - 11x + 30 = 0 \)

Explanation:
1. The roots are 2 and 3, so the new roots are 5 and 6.
2. Sum = 11, product = 30, so the equation is \( x^2 - 11x + 30 = 0 \).

Teacher's Note:
1. Alternatively replace x by x − 3 in the original equation.
2. Check the sum and product of the new roots.

Answer: (b) \( m = 6 \)

Explanation:
1. \( m^2 - 4m + 4 = (m - 2)^2 \).
2. \( D = (2 + m)^2 - 4(m - 2)^2 = 0 \), so \( 2 + m = \pm 2(m - 2) \).
3. Taking the + sign: \( m = 6 \). (The − sign gives \( m = \frac{2}{3} \), which is not an option.)

Teacher's Note:
1. Recognise the constant term as a perfect square.
2. Only one of the solutions appears among the options.

Answer: (a) \( x^2 - 2\sqrt{3}x + 5 = 0 \)

Explanation:
1. (a) \( D = 12 - 20 = -8 < 0 \).
2. (b) \( D = 72 - 64 = 8 > 0 \). (c) \( D = 12 + 20 > 0 \). (d) \( D = 72 + 72 > 0 \).
3. Only (a) has no real roots.

Teacher's Note:
1. No real roots means D < 0.
2. Compute the discriminant of each option.

Answer: (b) \( x^2 - 8x - 9 = 0 \)

Explanation:
1. Let the numbers be \( \alpha, \beta \): \( \alpha + \beta = 8 \), \( (\alpha - \beta)^2 = 100 \).
2. \( \alpha\beta = \frac{64 - 100}{4} = -9 \).
3. Equation: \( x^2 - 8x - 9 = 0 \) (roots 9 and −1).

Teacher's Note:
1. Use (α − β)² = (α + β)² − 4αβ.
2. The roots 9 and −1 check: sum 8, difference 10.

Answer: (d) \( a = -24 \)

Explanation:
1. \( \alpha + \beta = -5 \). Substituting \( \alpha = -5 - \beta \) in \( 2\alpha + 5\beta = -1 \): \( -10 + 3\beta = -1 \), so \( \beta = 3 \), \( \alpha = -8 \).
2. \( a = \alpha\beta = -24 \).

Teacher's Note:
1. Use the sum of the roots to eliminate one root.
2. Then find a from the product.

Answer: (c) 6

Explanation:
1. \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = p^2 - 32 = 4 \).
2. So \( p^2 = 36 \) and \( p = \pm 6 \). Among the options only 6 appears.

Teacher's Note:
1. Square the difference of the roots.
2. Only one of the two values of p is listed.

Answer: (b) 8

Explanation:
1. Let the roots be \( \alpha \) and \( 3\alpha \): \( 3\alpha^2 = 12 \), so \( \alpha = \pm 2 \).
2. \( p = -4\alpha = \mp 8 \), so \( p = \pm 8 \). Only 8 is listed.

Teacher's Note:
1. Use the product of the roots to find α first.
2. The sum of the roots then gives p.

Answer: (c) 2

Explanation:
1. \( (x - 1)(x - 2) = 0 \), so the roots are 1 and 2.
2. Among the options, 2 is a root.

Teacher's Note:
1. Factorise: two numbers with sum 3 and product 2.
2. Check the options by substitution.

Answer: (c) \( q = 9 \)

Explanation:
1. Sum of roots: \( 5 = \frac{2p}{3} \), so \( p = \frac{15}{2} \).
2. Product of roots: \( 6 = \frac{2q}{3} \), so \( q = 9 \).
3. Check (d): \( 6p + 2q = 45 + 18 = 63 \ne 27 \).

Teacher's Note:
1. Use sum and product, or substitute both roots.
2. Test every option with the values found.

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

Explanation:
1. Expanding, \( 4x^2 - 4x + 1 - 4x^2 + 5 = -4x + 6 = 0 \) is linear, so the Assertion is true.
2. In the Reason, \( 2x(x - 3) = 0 \) gives x = 0 or 3, so it is true.
3. The Reason does not explain the Assertion: option (b).

Teacher's Note:
1. Expand the Assertion before judging it.
2. Check that the two statements are not linked.

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

Explanation:
1. For equal roots \( D = 9k^2 - 144 = 0 \), so \( k^2 = 16 \) and \( k = \pm 4 \). The Assertion is true.
2. The Reason is the standard condition and explains the Assertion: option (a).

Teacher's Note:
1. Apply D = 0 to the given equation.
2. The Reason is the rule used.