CBSE Class 10 Mathematics Quadratic Equations MCQs Set 21

Answer: (c) \( (x + 2)^3 = x^3 - 4 \)

Explanation:
1. \( (x + 2)^3 = x^3 - 4 \) gives \( 6x^2 + 12x + 12 = 0 \), which is quadratic.
2. The first option has an \( x^3 \) term; the second simplifies to \( 25x - 10 = 0 \) (linear); the fourth simplifies to \( x + 12 = 0 \) (linear).

Teacher's Note:
1. Expand and simplify before deciding the degree.
2. When x³ or x² terms cancel, the degree drops.

Answer: (a) x² + x - 272 = 0

Explanation:
1. \( x^2 + (x + 1)^2 = 545 \) gives \( 2x^2 + 2x - 544 = 0 \).
2. Dividing by 2: \( x^2 + x - 272 = 0 \), with solutions \( x = 16 \) (integers 16 and 17).

Teacher's Note:
1. Expand (x + 1)² carefully.
2. Check: 256 + 289 = 545.

Answer: (d) x² - 5x - 300 = 0

Explanation:
1. The other side is \( (x - 5) \) cm. By Pythagoras: \( x^2 + (x - 5)^2 = 625 \).
2. \( 2x^2 - 10x - 600 = 0 \), i.e. \( x^2 - 5x - 300 = 0 \), with \( x = 20 \) (sides 20 and 15).

Teacher's Note:
1. Use a² + b² = c².
2. Check: 400 + 225 = 625.

Answer: (b) x² - 8x + 7 = 0

Explanation:
1. Father's age \( = x^2 \). One year ago: \( x^2 - 1 = 8(x - 1) \).
2. \( x^2 - 8x + 7 = 0 \), with \( x = 7 \) (or 1, rejected).

Teacher's Note:
1. Subtract 1 from each age.
2. Check: x = 7 gives father 49.

Answer: (b) p = -7, q = 0

Explanation:
1. Put \( x = 1 \): \( p + q + 7 = 0 \). Put \( x = -1 \): \( p - q + 7 = 0 \).
2. Adding: \( 2p + 14 = 0 \), so \( p = -7 \); then \( q = 0 \).

Teacher's Note:
1. A root satisfies the equation.
2. Add and subtract the two equations.

Answer: (d) \( \pm\frac{2}{3} \)

Explanation:
1. \( 27x^2 = 12 \) gives \( x^2 = \frac{4}{9} \).
2. So \( x = \pm\frac{2}{3} \).

Teacher's Note:
1. Take both square roots.
2. Check: 27 × 4/9 = 12.

Answer: (a) \( 0,\ -\frac{a + b}{3} \)

Explanation:
1. Expand: \( 9x^2 + 3(a + b)x + ab = ab \).
2. \( 9x^2 + 3(a + b)x = 0 \), so \( 3x(3x + a + b) = 0 \).
3. \( x = 0 \) or \( x = -\frac{a + b}{3} \).

Teacher's Note:
1. Do not cancel x; factor it out.
2. ab cancels on both sides.

Answer: (c) \( -\frac{2}{\sqrt{3}},\ -3\sqrt{3} \)

Explanation:
1. \( y = \frac{-11 \pm \sqrt{121 - 72}}{2\sqrt{3}} = \frac{-11 \pm 7}{2\sqrt{3}} \).
2. \( y = -\frac{2}{\sqrt{3}} \) or \( y = -\frac{18}{2\sqrt{3}} = -3\sqrt{3} \).

Teacher's Note:
1. Use the quadratic formula.
2. Check: product of roots is 6√3/√3 = 6.

Answer: (d) \( \frac{1}{3},\ -\frac{1}{4} \)

Explanation:
1. Multiply by 12: \( 12x^2 - x - 1 = 0 \).
2. \( 12x^2 - 4x + 3x - 1 = (3x - 1)(4x + 1) \).
3. \( x = \frac{1}{3} \) or \( x = -\frac{1}{4} \).

Teacher's Note:
1. Clear the fractions first.
2. Check: sum = 1/12 and product = -1/12.

Answer: (b) \( \frac{11}{5},\ \frac{5}{8} \)

Explanation:
1. Let \( t = \frac{x + 2}{2x - 3} \): \( 2t - \frac{9}{t} = 3 \) gives \( 2t^2 - 3t - 9 = 0 \), so \( t = 3 \) or \( t = -\frac{3}{2} \).
2. \( t = 3 \): \( x + 2 = 6x - 9 \), so \( x = \frac{11}{5} \).
3. \( t = -\frac{3}{2} \): \( 2x + 4 = -6x + 9 \), so \( x = \frac{5}{8} \).

Teacher's Note:
1. Substitution turns this into an ordinary quadratic in t.
2. Do not forget to convert back to x.

Answer: (a) \( \frac{1 + \sqrt{3}}{2},\ \frac{1 - \sqrt{3}}{2} \)

Explanation:
1. \( x^2 + (x - 1)^2 = 4x(x - 1) \) gives \( 2x^2 - 2x + 1 = 4x^2 - 4x \).
2. \( 2x^2 - 2x - 1 = 0 \).
3. \( x = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{1 \pm \sqrt{3}}{2} \).

Teacher's Note:
1. Multiply through by x(x - 1).
2. Quadratic formula gives the roots.

Answer: (c) \( -1,\ 3,\ 1 + \sqrt{2},\ 1 - \sqrt{2} \)

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

Teacher's Note:
1. Four roots in total because the equation has degree 4.
2. Solve each quadratic in x separately.

Answer: (a) -2 and 1

Explanation:
1. Let \( t = \frac{x}{x + 1} \): \( 2t^2 - 5t + 2 = 0 \), so \( t = 2 \) or \( t = \frac{1}{2} \).
2. \( t = 2 \): \( x = 2x + 2 \), so \( x = -2 \).
3. \( t = \frac{1}{2} \): \( 2x = x + 1 \), so \( x = 1 \).

Teacher's Note:
1. Use substitution.
2. Check x = 1: t = 1/2; x = -2: t = 2.

Answer: (d) 6 and -2

Explanation:
1. \( D = (p - 2)^2 - 16 = 0 \), so \( p - 2 = \pm 4 \).
2. \( p = 6 \) or \( p = -2 \).

Teacher's Note:
1. Equal roots mean D = 0.
2. Take both signs of the square root.

Answer: (c) x² + 5x - 500 = 0, x = 20

Explanation:
1. Litres used: \( \frac{400}{x} - \frac{400}{x + 5} = 4 \).
2. \( 100(x + 5) - 100x = x(x + 5) \), so \( x^2 + 5x - 500 = 0 \).
3. \( (x + 25)(x - 20) = 0 \), so \( x = 20 \) km per litre.

Teacher's Note:
1. Fuel used = distance ÷ mileage.
2. Reject the negative root.

Answer: (b) 20 km/h and 5 km/h

Explanation:
1. Let the speeds be \( u \) and \( v \). Then \( \frac{100}{u + v} + \frac{30}{u - v} = 6 \) and \( \frac{75}{u + v} + \frac{75}{u - v} = 8 \).
2. With \( a = \frac{1}{u + v} \), \( b = \frac{1}{u - v} \): \( a + b = \frac{8}{75} \) and \( 100a + 30b = 6 \), giving \( a = \frac{1}{25} \), \( b = \frac{1}{15} \).
3. So \( u + v = 25 \) and \( u - v = 15 \): \( u = 20 \), \( v = 5 \).

Teacher's Note:
1. Use reciprocals of the speeds as new variables.
2. Check: 100/25 + 30/15 = 4 + 2 = 6.

Answer: (c) 12

Explanation:
1. \( x^2 - 84 = 3(x + 8) \) gives \( x^2 - 3x - 108 = 0 \).
2. \( (x - 12)(x + 9) = 0 \); the natural number is 12.

Teacher's Note:
1. Natural numbers are positive.
2. Check: 144 - 84 = 60 = 3 × 20.

Answer: (a) 16

Explanation:
1. \( x + 84 = \frac{1600}{x} \) gives \( x^2 + 84x - 1600 = 0 \).
2. \( (x - 16)(x + 100) = 0 \); the natural number is 16.

Teacher's Note:
1. Reject the negative root.
2. Check: 16 + 84 = 100 = 1600/16.

Answer: (d) 1 - b

Explanation:
1. The equation is \( x^2 - ax - (a + b) = 0 \): \( \alpha + \beta = a \) and \( \alpha\beta = -(a + b) \).
2. \( (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = -a - b + a + 1 = 1 - b \).

Teacher's Note:
1. Expand the product first.
2. The a terms cancel.

Answer: (b) 20 m and 20 m (it is a square)

Explanation:
1. Let the length be \( x \); breadth \( = 40 - x \). Area: \( x(40 - x) = 400 \).
2. \( x^2 - 40x + 400 = (x - 20)^2 = 0 \), so \( x = 20 \).
3. Length = breadth = 20 m.

Teacher's Note:
1. Half the perimeter is length + breadth.
2. D = 0, so the only possibility is a square.

Answer: (b) 50

Explanation:
1. \( \frac{300}{x} - \frac{300}{x + 10} = 1 \).
2. \( 3000 = x^2 + 10x \), so \( x^2 + 10x - 3000 = 0 \) and \( (x - 50)(x + 60) = 0 \).
3. Number of students = 50.

Teacher's Note:
1. Share per student = total ÷ students.
2. Check: 6 each versus 5 each with 60 students.

Answer: (d) 10 and 6

Explanation:
1. Let the larger part be \( x \): \( 2x^2 - (16 - x)^2 = 164 \).
2. \( x^2 + 32x - 420 = 0 \), so \( (x - 10)(x + 42) = 0 \) and \( x = 10 \).
3. The parts are 10 and 6.

Teacher's Note:
1. The smaller part is 16 minus the larger.
2. Check: 200 - 36 = 164.

Answer: (a) 800 km/h

Explanation:
1. Let the usual speed be \( s \): \( \frac{1600}{s} - \frac{1600}{s + 400} = \frac{2}{3} \).
2. \( s^2 + 400s - 960000 = 0 \), so \( (s - 800)(s + 1200) = 0 \).
3. Usual speed = 800 km/h.

Teacher's Note:
1. 40 minutes = 2/3 hour.
2. Check: 2 h versus 1600/1200 = 4/3 h; difference 2/3 h.

Answer: (c) 16

Explanation:
1. \( \frac{80}{x} - \frac{80}{x + 4} = 1 \).
2. \( x^2 + 4x - 320 = 0 \), so \( (x - 16)(x + 20) = 0 \).
3. He bought 16 books.

Teacher's Note:
1. Price per book = total ÷ number.
2. Check: Rs. 5 each versus Rs. 4 each with 20 books.