CBSE Class 10 Mathematics Quadratic Equations MCQs Set 20

Answer: (b) If the discriminant of a quadratic equation is negative, then it has no real roots

Explanation:
1. A negative discriminant means \( b^2 - 4ac < 0 \), so the square root in the formula is not real and there is no real root.
2. Option (a) is false (the roots may be equal or non-real), (c) is false (\( -x^2 + 1 = 0 \) has real roots) and (d) is false (D = 0 gives two equal roots).

Teacher's Note:
1. Nature of roots depends only on the discriminant D = b² - 4ac.
2. D > 0: two distinct real roots; D = 0: equal real roots; D < 0: no real roots.

Answer: (b) -3, -4

Explanation:
1. The first candidate used the correct q: \( q = 2 \times 6 = 12 \).
2. The second used the correct p: \( -p = 2 + (-9) = -7 \), so \( p = 7 \).
3. The correct equation is \( x^2 + 7x + 12 = 0 \), i.e. \( (x + 3)(x + 4) = 0 \), so the roots are -3 and -4.

Teacher's Note:
1. A wrong p still gives the right product of roots; a wrong q still gives the right sum.
2. Always rebuild the equation before solving.

Answer: (d) \( \left(-\frac{b}{2a},\ -\frac{D}{4a}\right) \)

Explanation:
1. The vertex lies on the axis of symmetry \( x = -\frac{b}{2a} \).
2. Substituting gives \( y = c - \frac{b^2}{4a} = -\frac{b^2 - 4ac}{4a} = -\frac{D}{4a} \).
3. So the vertex is \( \left(-\frac{b}{2a}, -\frac{D}{4a}\right) \).

Teacher's Note:
1. The x-coordinate is the mean of the two roots.
2. Check with y = x²: a = 1, b = 0, D = 0 gives the vertex (0, 0).

Answer: (c) \( \frac{13}{\sqrt{7}},\ -\sqrt{7} \)

Explanation:
1. \( y = \frac{6 \pm \sqrt{36 + 4 \cdot 7 \cdot 13}}{2\sqrt{7}} = \frac{6 \pm 20}{2\sqrt{7}} \).
2. \( y = \frac{26}{2\sqrt{7}} = \frac{13}{\sqrt{7}} \) or \( y = \frac{-14}{2\sqrt{7}} = -\sqrt{7} \).

Teacher's Note:
1. Use the quadratic formula with a = √7, b = -6, c = -13√7.
2. Check the product of roots: (13/√7)(-√7) = -13 = c/a.

Answer: (d) 2b

Explanation:
1. \( x^2 - 2ax + a^2 - b^2 = (x - a)^2 - b^2 = (x - a - b)(x - a + b) \).
2. Roots: \( a + b \) and \( a - b \), so the difference is \( 2b \).

Teacher's Note:
1. Complete the square: (x - a)² - b².
2. Difference of roots = √D / |a| = √(4b²) = 2b.

Answer: (c) \( 2,\ -\frac{10}{9} \)

Explanation:
1. Write as \( x^2 - 2(1 + 3k)x + 7(3 + 2k) = 0 \). Equal roots need \( (1 + 3k)^2 - 7(3 + 2k) = 0 \).
2. \( 9k^2 + 6k + 1 - 21 - 14k = 9k^2 - 8k - 20 = 0 \).
3. \( (k - 2)(9k + 10) = 0 \), so \( k = 2 \) or \( k = -\frac{10}{9} \).

Teacher's Note:
1. Equal roots mean discriminant zero.
2. Arrange the terms in standard form first.

Answer: (a) -92

Explanation:
1. \( D = b^2 - 4ac = 2^2 - 4(-3)(-8) = 4 - 96 = -92 \).

Teacher's Note:
1. Keep the signs of a and c: (-3)(-8) = +24.
2. A negative D means no real roots.

Answer: (a) no real roots

Explanation:
1. \( D = 25 - 28 = -3 < 0 \).
2. So the equation has no real roots.

Teacher's Note:
1. Compute D before trying to solve.
2. If D < 0 the parabola does not meet the x-axis.

Answer: (d) b + c = 2a

Explanation:
1. The coefficients add up to \( (a - b) + (b - c) + (c - a) = 0 \), so \( x = 1 \) is a root.
2. Equal roots mean both are 1, so the product \( \frac{c - a}{a - b} = 1 \).
3. \( c - a = a - b \), so \( b + c = 2a \).

Teacher's Note:
1. When the coefficients add to zero, x = 1 is always a root.
2. Equal roots then force product = 1.

Answer: (b) p² < 4q

Explanation:
1. \( D = p^2 - 4q \).
2. No real roots need \( D < 0 \), i.e. \( p^2 < 4q \).

Teacher's Note:
1. Keep the inequality direction: D < 0.
2. Example: p = 2, q = 5 gives D = -16.

Answer: (b) \( \pm\frac{4}{3} \)

Explanation:
1. \( D = 9k^2 - 16 = 0 \).
2. \( k^2 = \frac{16}{9} \), so \( k = \pm\frac{4}{3} \).

Teacher's Note:
1. Equal roots: D = 0.
2. Both signs are valid.

Answer: (a) 12

Explanation:
1. Sum of roots \( = 8 \) and product \( = k \).
2. \( \alpha^2 + \beta^2 = 64 - 2k = 40 \), so \( k = 12 \).

Teacher's Note:
1. Use α² + β² = (α + β)² - 2αβ.
2. Check: x² - 8x + 12 has roots 2 and 6.

Answer: (d) Yes, if a ≠ b

Explanation:
1. \( D = 4(a + b)^2 - 8(a^2 + b^2) = -4(a - b)^2 \).
2. D is negative whenever \( a \ne b \), so the roots are non-real in that case (and equal when a = b).

Teacher's Note:
1. Simplify D first: 4(a + b)² - 8(a² + b²) = -4(a - b)².
2. A square is non-negative; the minus sign makes D ≤ 0.

Answer: (b) 34

Explanation:
1. Let the digits be x (tens) and y (units). \( 10x + y + 9 = 10y + x \) gives \( y = x + 1 \).
2. \( xy = x(x + 1) = 12 \), so \( x = 3 \) and \( y = 4 \).
3. The number is 34; check: 34 + 9 = 43.

Teacher's Note:
1. Form one equation from the interchange and one from the product.
2. Check by adding 9.

Answer: (b) 24

Explanation:
1. Let the numbers be \( n \) and \( n + 2 \): \( n^2 + (n + 2)^2 = 290 \).
2. \( n^2 + 2n - 143 = 0 \), so \( (n + 13)(n - 11) = 0 \) and \( n = 11 \).
3. Numbers 11 and 13, sum 24.

Teacher's Note:
1. Consecutive odd numbers differ by 2.
2. Reject the negative root.

Answer: (b) ax² + kbx + k²c = 0

Explanation:
1. Replace x by \( \frac{x}{k} \): \( a\frac{x^2}{k^2} + b\frac{x}{k} + c = 0 \).
2. Multiply by \( k^2 \): \( ax^2 + kbx + k^2c = 0 \).

Teacher's Note:
1. If y = kx then x = y/k.
2. Check with k = 2 and x² - 3x + 2 = 0 (roots 1, 2): new roots 2, 4 give x² - 6x + 8 = 0.

Answer: (c) 16

Explanation:
1. \( 4 + 2b + 12 = 0 \) gives \( b = -8 \).
2. \( x^2 - 8x + q = 0 \) has equal roots when \( 64 - 4q = 0 \), so \( q = 16 \).

Teacher's Note:
1. First find b from the known root.
2. Equal roots: D = 0.

Answer: (a) -3

Explanation:
1. \( D = 4(c + 1)^2 - 16(c + 4) = 0 \).
2. \( c^2 - 2c - 15 = 0 \), so \( (c - 5)(c + 3) = 0 \) and \( c = 5 \) or \( c = -3 \).
3. Among the given options only -3 appears, so the answer is -3.

Teacher's Note:
1. Both c = 5 and c = -3 make D = 0; only one of them is offered as an option.
2. Substitute c = -3: 4x² + 4x + 1 = (2x + 1)² confirms equal roots.

Answer: (a) Real and unequal

Explanation:
1. If a and c have opposite signs, then \( ac < 0 \).
2. \( D = b^2 - 4ac = b^2 + (\text{positive}) > 0 \), so the roots are real and unequal.

Teacher's Note:
1. -4ac becomes positive when ac is negative.
2. Example: x² - 4 = 0 has roots ±2.

Answer: (b) \( -\frac{27}{4} \)

Explanation:
1. \( \alpha + \beta = 5 \) and \( \alpha\beta = 4 \).
2. \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{5}{4} \) and \( 2\alpha\beta = 8 \).
3. \( \frac{5}{4} - 8 = -\frac{27}{4} \).

Teacher's Note:
1. Use sum and product of roots.
2. The roots are 1 and 4: 1 + 1/4 - 8 = -27/4.

Answer: (a) 49

Explanation:
1. Let the number be \( x^2 \) (with positive root x): \( x^2 - x = 42 \).
2. \( x^2 - x - 42 = (x - 7)(x + 6) = 0 \), so \( x = 7 \).
3. The number is \( 7^2 = 49 \).

Teacher's Note:
1. Treat the positive square root as the unknown.
2. Check: 49 - 7 = 42.

Answer: (c) \( 8\ \text{or}\ \frac{1}{8} \)

Explanation:
1. \( x + \frac{1}{x} = \frac{65}{8} \) gives \( 8x^2 - 65x + 8 = 0 \).
2. \( (8x - 1)(x - 8) = 0 \), so \( x = 8 \) or \( x = \frac{1}{8} \).

Teacher's Note:
1. Both roots are valid because they are reciprocals of each other.
2. Check: 8 + 1/8 = 65/8.

Answer: (d) 7, 49

Explanation:
1. Let the son's present age be x. Father's present age \( = x^2 \).
2. One year ago: \( x^2 - 1 = 8(x - 1) \), so \( x + 1 = 8 \) and \( x = 7 \).
3. Son 7 years, father 49 years.

Teacher's Note:
1. Subtract 1 from both ages for the earlier condition.
2. Cancel the common factor (x - 1).

Answer: (a) 3, 4, 5

Explanation:
1. Let the integers be \( n, n + 1, n + 2 \).
2. \( n^2 + (n + 1)^2 + (n + 2)^2 = 3n^2 + 6n + 5 = 50 \), so \( n^2 + 2n - 15 = 0 \).
3. \( n = 3 \), giving 3, 4, 5 (check: 9 + 16 + 25 = 50).

Teacher's Note:
1. Reject the negative root n = -5 since the integers are positive.
2. Check by adding the squares.

Answer: (b) 24

Explanation:
1. Let the digits be x and y: \( 10x + y = 4(x + y) = 3xy \).
2. From \( 10x + y = 4x + 4y \): \( y = 2x \). From \( 4(x + y) = 3xy \): \( 12x = 6x^2 \), so \( x = 2 \), \( y = 4 \).
3. The number is 24: check \( 4 \times 6 = 24 \) and \( 3 \times 8 = 24 \).

Teacher's Note:
1. Use two equations and substitute y = 2x.
2. Check both conditions.