CBSE Class 10 Mathematics Quadratic Equations MCQs Set 32

Answer: (c) unequal but of opposite signs

Explanation:
1. \( x^{2} - 5x - 6 = (x - 6)(x + 1) \), so the roots are 6 and \( -1 \).
2. The roots are unequal and have opposite signs (the product of roots is \( -6 \), which is negative).

Teacher's Note:
1. The sign of the product of roots tells whether the roots have the same or opposite signs.
2. A negative product means opposite signs.

Answer: (a) \( x^2 - 6x - 3 = 0 \)

Explanation:
1. Irrational roots of a rational quadratic occur in conjugate pairs, so the other root is 3 - 2√3.
2. Sum = 6 and product = \( 9 - 12 = -3 \).
3. The equation is \( x^{2} - 6x - 3 = 0 \).

Teacher's Note:
1. Use \( x^{2} \) - (sum)x + product = 0.
2. Conjugate surd is the other root.

Answer: (c) a = c

Explanation:
1. Product of roots = \( c/a \).
2. Reciprocal roots have product 1, so \( c/a = 1 \) and \( a = c \).

Teacher's Note:
1. Product of roots is \( c/a \).
2. Roots \( r \) and \( 1/r \) multiply to 1.

Answer: (a) 0

Explanation:
1. Discriminant \( D = 4a^{2}(b^{2} + c^{2}) - 4(a^{2} + b^{2})(a^{2} + c^{2}) \).
2. Expanding: \( D = 4[a^{2}b^{2} + a^{2}c^{2} - a^{4} - a^{2}c^{2} - a^{2}b^{2} - b^{2}c^{2}] = -4(a^{4} + b^{2}c^{2}) \).
3. Since a, \( b, c \) are non-zero, \( D \lt 0 \), so there is no real root.

Teacher's Note:
1. Compute the discriminant and expand carefully.
2. A negative discriminant means no real roots.

Answer: (b) 4

Explanation:
1. Let the roots be α and β: \( \alpha + \beta = 4 \) and \( \alpha\beta = 1 \).
2. √(p/q) + √(q/p) = (p + q)/√(pq) = (α + β)/√(αβ) (ratios only matter).
3. = \( 4/1 = 4 \).

Teacher's Note:
1. Write the expression over a common denominator.
2. Use the sum and product of the roots.

Answer: (c) The roots of one equation are the reciprocals of the roots of the other.

Explanation:
1. If α is a root of ax² + bx \( + c = 0 \), then \( a\alpha^{2} + b\alpha + c = 0 \).
2. Dividing by α²: \( a + b(1/\alpha) + c(1/\alpha)^{2} = 0 \), so 1/α is a root of cx² + bx \( + a = 0 \).

Teacher's Note:
1. Substitute \( x = 1/\alpha \) in the second equation.
2. Reversing the order of coefficients reciprocates the roots.

Answer: (c) \( -p = q^{1/4}\left[1 + q^{1/2}\right] \)

Explanation:
1. Let the roots be α and α³. Then \( \alpha + \alpha^{3} = -p \) and \( \alpha\cdot \alpha^{3} = \alpha^{4} = q \).
2. So \( \alpha = q^{1/4} \) and \( \alpha^{3} = q^{3/4} = q^{1/4}\cdot q^{1/2} \).
3. \( -p = q^{1/4} + q^{3/4} = q^{1/4}[1 + q^{1/2}] \).

Teacher's Note:
1. Write both roots in terms of one letter.
2. Use sum and product of roots.

Answer: (c) -1

Explanation:
1. Subtract the equations: \( (a - b)x + (b - a) = 0 \), so \( (a - b)(x - 1) = 0 \) and \( x = 1 \) (as \( a \ne b \)).
2. Put \( x = 1 \): \( 1 + a + b = 0 \), so \( a + b = -1 \).

Teacher's Note:
1. Subtract the two equations to eliminate \( x^{2} \).
2. The common root must satisfy both.

Answer: (c) 4

Explanation:
1. Let \( t = |x - 2| \): \( t^{2} + t - 2 = 0 \), so \( (t + 2)(t - 1) = 0 \).
2. \( t = -2 \) is impossible; \( t = 1 \) gives \( x - 2 = \pm 1 \), so \( x = 3 \) or \( x = 1 \).
3. Sum = 4.

Teacher's Note:
1. Substitute \( t = |x - 2| \) and reject negative values of \( t \).
2. Each valid \( t \) gives two values of \( x \).

Answer: (c) real

Explanation:
1. The equation is \( 3x^{2} - 2(a + b + c)x \) + (ab + bc + ca) = 0.
2. D = 4[(a + b + c)² - 3(ab + bc + ca)] = 4[a² + b² + c² - ab - bc - ca] = \( 2[(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \ge 0 \).
3. The discriminant is never negative, so the roots are always real (not necessarily positive or negative).

Teacher's Note:
1. Expand to standard form and test the discriminant.
2. \( a^{2} + b^{2} + c^{2} \) - ab - bc - ca is a sum of squares.

Answer: (a) 4

Explanation:
1. Let \( t = |x| \): \( t^{2} - 3t + 2 = 0 \), so \( t = 1 \) or \( t = 2 \).
2. \( |x| = 1 \) gives \( x = \pm 1 \) and \( |x| = 2 \) gives \( x = \pm 2 \).
3. There are 4 solutions.

Teacher's Note:
1. Substitute \( t = |x| \).
2. Each positive value of \( |x| \) gives two values of \( x \).

Answer: (a) no root

Explanation:
1. Here \( x \ne 1 \).
2. Add \( 2/(x - 1) \) to both sides: \( x = 1 \).
3. But \( x = 1 \) is excluded, so the equation has no root.

Teacher's Note:
1. Note the values excluded from the domain before solving.
2. Check the root against the exclusion.

Answer: (c) a and b

Explanation:
1. Since α, β are the roots of \( (x - a)(x - b) - c = 0 \), we have \( (x - \alpha)(x - \beta) = (x - a)(x - b) - c \).
2. Then \( (x - \alpha)(x - \beta) + c = (x - a)(x - b) \).
3. Its roots are a and \( b \).

Teacher's Note:
1. Two quadratics with the same roots and leading coefficient are identical.
2. Rewrite \( (x - \alpha)(x - \beta) \) using the first equation.

Answer: (a) no solution

Explanation:
1. The equation needs \( x \ge 1 \).
2. For \( x \ge 1, \sqrt{x + 1} - \sqrt{x - 1} = 2/(\sqrt{x + 1} + \sqrt{x - 1}) \le 2/(\sqrt{2}) = \sqrt{2} \).
3. But \( \sqrt{4x - 1} \ge \sqrt{3} \gt \sqrt{2} \), so the two sides are never equal.

Teacher's Note:
1. Find the domain first.
2. Compare the greatest value of the left side with the least value of the right side.

Answer: (a) a < 2

Explanation:
1. \( D = 4a^{2} - 4(a^{2} + a - 3) = 4(3 - a) \ge 0 \), so \( a \le 3 \).
2. Roots are \( a \pm \sqrt{3 - a} \). The larger root \( a + \sqrt{3 - a} \) < 3 gives \( \sqrt{3 - a} \) < \( 3 - a \).
3. Put \( t = 3 - a \) > 0: \( \sqrt{t} \) < \( t \) means \( t \) > 1, i.e. a < 2.

Teacher's Note:
1. The larger root must be less than 3.
2. Use \( t = 3 - a \) to simplify.

Answer: (a) \( p^3 - q(3p - 1) + q^2 = 0 \)

Explanation:
1. Let the roots be α and α². Then \( \alpha + \alpha^{2} = -p \) and \( \alpha^{3} = q \).
2. Cube the sum: \( (\alpha + \alpha^{2})^{3} = \alpha^{3} + \alpha^{6} + 3\alpha^{4}(1 + \alpha)\cdot ... = q + q^{2} + 3q(\alpha + \alpha^{2}) = q + q^{2} \) - 3pq.
3. So \( -p^{3} = q + q^{2} \) - 3pq, i.e. \( p^{3} - q(3p - 1) + q^{2} = 0 \).

Teacher's Note:
1. Use \( \alpha + \alpha^{2} = -p \) and \( \alpha^{3} = q \).
2. Expand the cube of the sum: \( (\alpha + \alpha^{2})^{3} = \alpha^{3} + \alpha^{6} + 3\alpha^{3}(\alpha + \alpha^{2}) \).

Answer: (b) 6

Explanation:
1. Put \( x = 1 \) in the first: \( 1 + a + 2 = 0 \), so \( a = -3 \).
2. Put \( x = 1 \) in the second: \( 1 + 1 + b = 0 \), so \( b = -2 \).
3. ab = 6.

Teacher's Note:
1. Substitute the common root into each equation separately.
2. Then multiply a and \( b \).

Answer: (a) \( \frac{216}{x} \) hours

Explanation:
1. Time = distance / speed = \( 216/x \) hours.

Teacher's Note:
1. Use time = distance ÷ speed.
2. The car covers 216 km by road.

Answer: (b) \( \frac{208}{16 + x} \) hours

Explanation:
1. The train's speed is \( (x + 16) \) km/h and it covers 208 km.
2. Time = \( 208/(x + 16) \) hours.

Teacher's Note:
1. Add 16 to the car's speed to get the train's speed.
2. Use time = distance ÷ speed.

Answer: (c) 36 km/h

Explanation:
1. \( 216/x - 208/(x + 16) = 2 \).
2. \( 216(x + 16) - 208x = 2x(x + 16) \), so \( 8x + 3456 = 2x^{2} + 32x \).
3. \( x^{2} + 12x - 1728 = 0 \), so \( (x - 36)(x + 48) = 0 \) and \( x = 36 \) (speed cannot be negative).

Teacher's Note:
1. Form the equation from the time difference.
2. Reject the negative root.

Answer: (a) \( -4\sqrt{3} \) and \( \frac{2}{\sqrt{3}} \)

Explanation:
1. \( D = 100 + 4(\sqrt{3})(8\sqrt{3}) = 100 + 96 = 196 \), so \( \sqrt{D} = 14 \).
2. \( x = (-10 \pm 14)/(2\sqrt{3}) = 2/\sqrt{3} \) or \( -24/(2\sqrt{3}) = -12/\sqrt{3} = -4\sqrt{3} \).

Teacher's Note:
1. Apply the quadratic formula.
2. Rationalise the roots to check them.

Answer: (a) \( -\frac{3}{2} \) and \( \frac{3}{5p} \)

Explanation:
1. Group the terms: 5px(2x \( + 3) - 3(2x + 3 \)) = (5px \( - 3)(2x + 3 \)).
2. Roots: \( x = 3/(5p) \) and \( x = -3/2 \).

Teacher's Note:
1. Group the terms with \( 2x + 3 \) as the common factor.
2. Set each factor equal to zero.

Answer: (b) 3

Explanation:
1. \( D = 4(2k - 3)^{2} - 4(k - 2)(5k - 6) = 0 \).
2. \( (4k^{2} - 12k + 9) - (5k^{2} - 16k + 12) = 0 \), so \( -k^{2} + 4k - 3 = 0 \), i.e. \( (k - 1)(k - 3) = 0 \).
3. \( k = 1 \) or 3. Since \( k \) > \( 2, k = 3 \).

Teacher's Note:
1. Equal roots need \( D = 0 \).
2. Use the condition \( k \) > 2 to choose between the two values.

Answer: (a) 0

Explanation:
1. Let \( s = \sec \theta \): \( 8s^{2} - 6s + 1 = 0 \), so \( (4s - 1)(2s - 1) = 0 \) and \( s = 1/4 \) or 1/2.
2. But \( |\sec \theta| \ge 1 \), so neither value is possible.
3. The equation has no real root.

Teacher's Note:
1. Solve for \( \sec \theta \) first.
2. Remember the range of \( \sec \theta \).

Answer: (a) \( 2\sqrt{2} \)

Explanation:
1. \( D = k^{2} - 8 = 0 \), so \( k = \pm 2\sqrt{2} \).
2. \( k \) positive gives \( k = 2\sqrt{2} \).

Teacher's Note:
1. Equal roots need \( D = 0 \).
2. Pick the positive value.