CBSE Class 10 Mathematics Quadratic Equations MCQs Set 31

Answer: (c) 1, −8

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

Teacher's Note:
1. A substitution such as \( t = x^{1/3} \) turns this into an ordinary quadratic.
2. Cube the values of \( t \) to return to \( x \).

Answer: (d) 1

Explanation:
1. Put \( t = 2^x \). Then \( t^{2} + t - 6 = 0 \), so \( (t + 3)(t - 2) = 0 \).
2. \( t = -3 \) is impossible because \( 2^x \) is always positive. So \( 2^x = 2 \) and \( x = 1 \).

Teacher's Note:
1. An exponential expression is always positive, so reject negative values of \( t \).
2. Check: \( 4 + 2 = 6 \).

Answer: (c) no solution

Explanation:
1. Both square roots need \( x \ge 1 \) to be real, and each root is at least 0.
2. A sum of two non-negative numbers is zero only if both are zero, i.e. \( x + 1 = 0 \) and \( x - 1 = 0 \) together, which is impossible.

Teacher's Note:
1. Square roots are never negative, so their sum cannot vanish unless each term vanishes.
2. Squaring blindly would introduce false solutions.

Answer: (c) 0, 1

Explanation:
1. Write \( \sqrt{3x+1} = 1 + \sqrt{x} \) and square: \( 3x + 1 = 1 + 2\sqrt{x} + x \), so \( 2x = 2\sqrt{x} \).
2. Hence \( \sqrt{x}(\sqrt{x} - 1) = 0 \), so \( x = 0 \) or \( x = 1 \).
3. Check: \( x = 0 \) gives \( 1 - 1 = 0 \) ✓. \( x = 1 \) gives \( 2 - 1 = 1 \) ✓. Both are roots.

Teacher's Note:
1. After squaring a radical equation, always substitute the answers back.
2. Here both values survive the check.

Answer: (d) 3

Explanation:
1. Let the value be \( y \). Then \( y = \sqrt{6 + y} \), so \( y^{2} = 6 + y \).
2. \( y^{2} - y - 6 = 0 \) gives \( (y - 3)(y + 2) = 0 \). Since \( y \gt 0, y = 3 \).

Teacher's Note:
1. An infinite nested radical repeats itself, so it satisfies \( y = \sqrt{6 + y} \).
2. Reject the negative root.

Answer: (b) p = 1 or 0

Explanation:
1. Sum of roots: \( p + q = -p \), so \( q = -2p \). Product of roots: pq = \( q \).
2. pq = \( q \) gives \( q(p - 1) = 0 \), so \( q = 0 \) or \( p = 1 \).
3. If \( q = 0 \), then \( -2p = 0 \) and \( p = 0 \). If \( p = 1 \), then \( q = -2 \).
4. So \( p = 1 \) or \( p = 0 \).

Teacher's Note:
1. Use sum = \( -b/a \) and product = \( c/a \), and do not divide by \( q \) because \( q \) may be zero.
2. Check both cases in the original equation.

Answer: (a) 8

Explanation:
1. \( \sqrt{2x + 9} = 13 - x \). Squaring: \( 2x + 9 = 169 - 26x + x^{2} \), so \( x^{2} - 28x + 160 = 0 \).
2. \( (x - 8)(x - 20) = 0 \), so \( x = 8 \) or 20.
3. Check \( x = 8 \): \( \sqrt{25} + 8 = 13 \) ✓. \( x = 20 \): \( \sqrt{49} + 20 = 27 \ne 13 \), rejected.

Teacher's Note:
1. Isolate the radical before squaring.
2. Extraneous roots are common, so always test the values.

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

Explanation:
1. Let \( D = x + 1/(x + 1/(x + ...)) \). The pattern repeats, so \( D = x + 1/D \).
2. Also \( x = 1 + 1/D \), so \( D = 1/(x - 1) \).
3. Then \( 1/(x - 1) = x + (x - 1) = 2x - 1 \), so \( (x - 1)(2x - 1) = 1 \) and \( 2x^{2} - 3x = 0 \).
4. \( x = 0 \) is impossible because \( x \gt 1 \), so \( x = 3/2 \).

Teacher's Note:
1. In a repeating continued fraction, the tail equals the whole denominator.
2. Check: \( x = 3/2 \) gives \( D = 2 \) and \( 1 + 1/2 = 3/2 \).

Answer: (a) real and distinct

Explanation:
1. \( D = a^{2} - 4(1)(-4) = a^{2} + 16 \).
2. \( a^{2} + 16 \gt 0 \) for every real a, so the roots are real and distinct.

Teacher's Note:
1. When a and \( c \) have opposite signs, ac is negative and \( D = b^{2} \) − 4ac is always positive.
2. Option \( (d) \) 'real' is true but less precise; the roots are always unequal.

Answer: (c) k = 0, 3

Explanation:
1. A perfect square needs \( D = 0 \): \( (2k + 4)^{2} - 4(4 - k)(8k + 1) = 0 \).
2. \( (4 - k)(8k + 1) = -8k^{2} + 31k + 4 \), so \( 4k^{2} + 16k + 16 + 32k^{2} - 124k - 16 = 0 \).
3. \( 36k^{2} - 108k = 0 \), so \( k = 0 \) or \( k = 3 \).

Teacher's Note:
1. A quadratic is a perfect square exactly when its discriminant is zero.
2. Check \( k = 0 \): \( 4x^{2} + 4x + 1 = (2x + 1)^{2} \).

Answer: (c) −½

Explanation:
1. Rewrite: \( x^{2} - 2(k + 1)x + k^{2} = 0 \). For equal roots \( D = 0 \).
2. \( 4(k + 1)^{2} - 4k^{2} = 0 \) gives \( 4(2k + 1) = 0 \), so \( k = -\frac{1}{2} \).

Teacher's Note:
1. Collect the \( x \) terms first so that \( b \) is read correctly.
2. Check: \( k = -\frac{1}{2} \) gives \( x^{2} - x + \frac{1}{4} = (x - \frac{1}{2})^{2} \).

Answer: (c) 9, 5

Explanation:
1. Write it as \( x^{2} - (3k - 1)x + (2k^{2} + 2k - 11) = 0 \) and put \( D = 0 \).
2. \( (3k - 1)^{2} - 4(2k^{2} + 2k - 11) = 9k^{2} - 6k + 1 - 8k^{2} - 8k + 44 = k^{2} - 14k + 45 = 0 \).
3. \( (k - 9)(k - 5) = 0 \), so \( k = 9 \) or 5.

Teacher's Note:
1. Bring all terms to one side before reading the constant term.
2. Mind the 11 on the right side.

Answer: (c) Rs. 20

Explanation:
1. Let the original price be Rs. \( p \). Books bought: \( 300/(p - 5) - 300/p = 5 \).
2. \( 300 \times 5 / (p(p - 5)) = 5 \), so \( p^{2} - 5p - 300 = 0 \).
3. \( (p - 20)(p + 15) = 0 \), so \( p = 20 \) (negative rejected).

Teacher's Note:
1. Number of books = total money ÷ price.
2. Check: \( 300/15 = 20 \) books and \( 300/20 = 15 \) books, a difference of 5.

Answer: (a) 11, 13

Explanation:
1. Let the integers be \( x \) and \( x + 2 \): \( x^{2} + (x + 2)^{2} = 290 \), so \( x^{2} + 2x - 143 = 0 \).
2. \( (x + 13)(x - 11) = 0 \), so \( x = 11 \) (positive). The integers are 11 and 13.

Teacher's Note:
1. Consecutive odd integers differ by 2.
2. Check: \( 121 + 169 = 290 \).

Answer: (c) 12, 8

Explanation:
1. Let the numbers be a (larger) and \( b \): \( a^{2} + b^{2} = 208 \) and \( a^{2} = 18b \).
2. \( 18b + b^{2} = 208 \), so \( b^{2} + 18b - 208 = 0 \) and \( (b - 8)(b + 26) = 0 \), so \( b = 8 \).
3. \( a^{2} = 144 \), so \( a = 12 \).

Teacher's Note:
1. Substitute \( a^{2} \) from the second condition into the first to get a quadratic in \( b \).
2. Check: \( 144 + 64 = 208 \).

Answer: (c) 8, 15, 17

Explanation:
1. Let the hypotenuse be \( h \). Base = \( h - 2 \) and altitude = \( (h - 1)/2 \).
2. Pythagoras: \( (h - 2)^{2} + ((h - 1)/2)^{2} = h^{2} \), so \( 4(h - 2)^{2} + (h - 1)^{2} = 4h^{2} \).
3. \( h^{2} - 18h + 17 = 0 \), so h = 17 (h = 1 is impossible). Base = 15, altitude = 8.

Teacher's Note:
1. Express all three sides in one variable and apply Pythagoras.
2. Check: \( 8^{2} + 15^{2} = 289 = 17^{2} \).

Answer: (b) 16

Explanation:
1. Let \( n \) be the number of books: \( 80/n - 80/(n + 4) = 1 \).
2. \( 320 = n(n + 4) \), so \( n^{2} + 4n - 320 = 0 \) and \( (n + 20)(n - 16) = 0 \).
3. \( n = 16 \).

Teacher's Note:
1. Cost per book = total cost ÷ number of books.
2. Check: \( 80/16 = 5 \) and \( 80/20 = 4 \).

Answer: (d) 16 cm, 20 cm

Explanation:
1. \( x^{2} + (x + 4)^{2} = 656 \), so \( 2x^{2} + 8x - 640 = 0 \) and \( x^{2} + 4x - 320 = 0 \).
2. \( (x + 20)(x - 16) = 0 \), so \( x = 16 \). The sides are 16 cm and 20 cm.

Teacher's Note:
1. Reject the negative root because a length cannot be negative.
2. Check: \( 256 + 400 = 656 \).

Answer: (c) −5

Explanation:
1. Sum of roots = \( 6 + (-1) = 5 = -p \), so \( p = -5 \).
2. Check the product: \( 6 \times (-1) = -6 \) ✓.

Teacher's Note:
1. Use sum of roots = \( -b/a \).
2. The product of roots confirms the constant term.

Answer: (a) x² − 2x − 4 = 0

Explanation:
1. Sum of roots = 2 and product = \( 1 - 5 = -4 \).
2. The equation is \( x^{2} \) − (sum)x + product = 0, i.e. \( x^{2} - 2x - 4 = 0 \).

Teacher's Note:
1. The required equation is \( x^{2} \) − (sum of roots)x + (product of roots) = 0.
2. Mind the signs when the product is negative.

Answer: (a) (1, 2)

Explanation:
1. Let \( u = 1/x \) and \( v = 1/y \): \( u + v = 3/2 \) and \( u^{2} + v^{2} = 5/4 \).
2. uv = \( ((3/2)^{2} - 5/4)/2 = 1/2 \), so \( u \) and \( v \) are the roots of \( t^{2} - (3/2)t + 1/2 = 0 \), i.e. \( t = 1 \) and ½.
3. Hence \( (x, y) = (1, 2) \) or (2, 1). Only (1, 2) is among the options.

Teacher's Note:
1. Use \( (u + v)^{2} = u^{2} + v^{2} \) + 2uv to find uv, then form a quadratic.
2. Check (1, 2): \( 1 + \frac{1}{2} = 3/2 \) and \( 1 + \frac{1}{4} = 5/4 \).

Answer: (b) \( \frac{1}{2}, 5 \)

Explanation:
1. Sum of roots = \( 5/a = 10 \), so \( a = \frac{1}{2} \).
2. Product of roots = \( c/a = 10 \), so \( c = 10a = 5 \).

Teacher's Note:
1. Sum of roots = \( -b/a \) and product = \( c/a \).
2. Check: \( \frac{1}{2}x^{2} - 5x + 5 = 0 \) has sum 10 and product 10.

Answer: (c) 2b² = 9ac

Explanation:
1. Let the roots be \( r \) and \( 2r \). Sum: \( 3r = -b/a \), so \( r = -b/(3a) \). Product: \( 2r^{2} = c/a \).
2. \( 2b^{2}/(9a^{2}) = c/a \), so \( 2b^{2} \) = 9ac.

Teacher's Note:
1. Name the roots \( r \) and \( 2r \) and use both sum and product.
2. Test with \( x^{2} - 3x + 2 = 0 \) (roots 1, 2): \( 2(9) = 18 = 9(2) \) ✓.