CBSE Class 10 Mathematics Quadratic Equations MCQs Set 25

Answer: (b) \( p = -6 \) and the other root \( 1 \)

Explanation:
1. \( 2(4) + 2p + 4 = 0 \) gives \( p = -6 \).
2. The equation is \( 2x^2 - 6x + 4 = 0 \), i.e. \( x^2 - 3x + 2 = (x - 1)(x - 2) \), so the other root is 1.

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (d) \( p = 7 \) and the other root \( -\dfrac{2}{3} \)

Explanation:
1. \( 27 - 3p - 6 = 0 \) gives \( p = 7 \).
2. Product of roots \( = -2 \), so the other root is \( -\dfrac{2}{3} \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (c) \( \pm 2\sqrt{2} \)

Explanation:
1. \( D = 64 - 8p^2 = 0 \), so \( p^2 = 8 \) and \( p = \pm 2\sqrt{2} \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (d) \( \dfrac{41}{4} \)

Explanation:
1. \( \alpha + \beta = -\dfrac{5}{2} \) and \( \alpha\beta = -2 \).
2. \( \alpha^2 + \beta^2 = \dfrac{25}{4} + 4 = \dfrac{41}{4} \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (a) \( -\dfrac{41}{8} \)

Explanation:
1. \( \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \dfrac{41/4}{-2} = -\dfrac{41}{8} \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (c) \( \dfrac{73}{4} \)

Explanation:
1. \( \alpha^2 + \beta^2 = \dfrac{41}{4} \) and \( 4\alpha\beta = -8 \).
2. \( \dfrac{41}{4} + 8 = \dfrac{73}{4} \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (d) \( 198 \)

Explanation:
1. \( \alpha^3 + \beta^3 = 6^3 - 3(1)(6) = 216 - 18 = 198 \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (c) 35

Explanation:
1. \( \alpha + \beta = 5 \) and \( \alpha\beta = 6 \).
2. \( \alpha^3 + \beta^3 = 125 - 3(6)(5) = 35 \). (Check: roots 2 and 3 give 8 + 27 = 35.)

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

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

Explanation:
1. Replace \( x \) by \( \dfrac{1}{x} \): \( \dfrac{2}{x^2} + \dfrac{1}{x} - 1 = 0 \), i.e. \( 2 + x - x^2 = 0 \).
2. So the equation is \( x^2 - x - 2 = 0 \). (Check: the roots \( \dfrac{1}{2}, -1 \) become \( 2, -1 \).)

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (a) \( x^2 - 4x + 4 = 0 \)

Explanation:
1. \( \alpha^2 - 3\alpha = -5 \), so \( \alpha^2 - 3\alpha + 7 = 2 \); similarly the second is 2.
2. Both roots are 2: \( x^2 - 4x + 4 = 0 \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (b) \( -1 \)

Explanation:
1. Product \( = \dfrac{2m - 1}{m} = 3 \), so \( 2m - 1 = 3m \) and \( m = -1 \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (b) \( \dfrac{b^2 - 2ac}{a^2} \)

Explanation:
1. \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \dfrac{b^2}{a^2} - \dfrac{2c}{a} = \dfrac{b^2 - 2ac}{a^2} \).

Teacher's Note:
1. Sum of roots = -b/a and product of roots = c/a.
2. Always bring the equation to the form ax² + bx + c = 0 first.

Answer: (a) \( 4\alpha^3 - 3\alpha \)

Explanation:
1. Let the other root be \( \beta \); then \( \beta = -\dfrac{1}{2} - \alpha \) (sum of roots).
2. From \( 4\alpha^2 = 1 - 2\alpha \): \( 4\alpha^3 = \alpha - 2\alpha^2 = \alpha - \dfrac{1 - 2\alpha}{2} = 2\alpha - \dfrac{1}{2} \).
3. So \( 4\alpha^3 - 3\alpha = -\alpha - \dfrac{1}{2} = \beta \).

Teacher's Note:
1. Use the sum of roots and reduce α³ using the given equation.
2. Then substitute to express the other root in terms of α.

Answer: (c) \( 1, 5 \)

Explanation:
1. Multiply by \( x \): \( x^2 - 6x + 5 = 0 \).
2. \( (x - 1)(x - 5) = 0 \), so \( x = 1 \) or \( 5 \).

Teacher's Note:
1. Clear fractions by multiplying by the common denominator.
2. Reject any value that makes a denominator zero.

Answer: (b) \( 1 \pm \dfrac{1}{\sqrt{3}} \)

Explanation:
1. \( \dfrac{(x - 2) - x}{x(x - 2)} = 3 \) gives \( -2 = 3x^2 - 6x \).
2. \( 3x^2 - 6x + 2 = 0 \), so \( x = \dfrac{6 \pm \sqrt{12}}{6} = 1 \pm \dfrac{1}{\sqrt{3}} \).

Teacher's Note:
1. Clear fractions by multiplying by the common denominator.
2. Reject any value that makes a denominator zero.

Answer: (a) \( 1, 2 \)

Explanation:
1. \( \dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30} \) gives \( \dfrac{-11}{x^2 - 3x - 28} = \dfrac{11}{30} \).
2. \( x^2 - 3x - 28 = -30 \), so \( x^2 - 3x + 2 = 0 \) and \( x = 1 \) or \( 2 \).

Teacher's Note:
1. Clear fractions by multiplying by the common denominator.
2. Reject any value that makes a denominator zero.

Answer: (a) \( 3 \) or \( \dfrac{1}{3} \)

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

Teacher's Note:
1. Clear fractions by multiplying by the common denominator.
2. Reject any value that makes a denominator zero.

Answer: (d) \( 21 \) and \( 24 \)

Explanation:
1. Let the numbers be \( x \) and \( x + 3 \): \( x(x + 3) = 504 \).
2. \( x^2 + 3x - 504 = 0 \), i.e. \( (x + 24)(x - 21) = 0 \), so the positive value is \( x = 21 \).
3. The numbers are 21 and 24.

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.

Answer: (b) \( 16 \) m

Explanation:
1. \( x(x + 4) = 192 \) gives \( x^2 + 4x - 192 = 0 \).
2. \( (x + 16)(x - 12) = 0 \), so the breadth is 12 m and the length is 16 m.

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.

Answer: (c) \( 18 \) cm

Explanation:
1. Let the altitude be \( x \) cm: \( \dfrac{1}{2}x(x + 6) = 108 \), so \( x^2 + 6x - 216 = 0 \).
2. \( (x + 18)(x - 12) = 0 \), so the altitude is 12 cm and the base is 18 cm.

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.

Answer: (d) \( 9 \) years

Explanation:
1. \( (x - 3)(x + 5) = 84 \) gives \( x^2 + 2x - 99 = 0 \).
2. \( (x + 11)(x - 9) = 0 \), so the age is 9 years.

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.

Answer: (b) \( 40 \) km/h

Explanation:
1. \( \dfrac{360}{x} - \dfrac{360}{x + 5} = 1 \) gives \( x^2 + 5x - 1800 = 0 \).
2. \( (x + 45)(x - 40) = 0 \), so the speed is 40 km/h.

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.

Answer: (d) \( 600 \) km/h

Explanation:
1. \( \dfrac{1500}{x - 100} - \dfrac{1500}{x} = \dfrac{1}{2} \) gives \( x^2 - 100x - 300000 = 0 \).
2. \( (x - 600)(x + 500) = 0 \), so the usual speed is 600 km/h. (Check: 1500/500 - 1500/600 = 3 - 2.5 = 0.5.)

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.

Answer: (a) \( 25 \) hours

Explanation:
1. Let the larger tap take \( x \) hours: \( \dfrac{1}{x} + \dfrac{1}{x + 10} = \dfrac{8}{75} \).
2. \( 4x^2 - 35x - 375 = 0 \), i.e. \( (4x + 25)(x - 15) = 0 \), so \( x = 15 \).
3. The smaller tap takes \( 15 + 10 = 25 \) hours.

Teacher's Note:
1. Define the unknown, form the quadratic equation, then solve it.
2. Reject roots that are impossible, such as a negative length or age.