CBSE Class 10 Mathematics Quadratic Equations MCQs Set 22

Answer: (b) \( \sqrt{3}x^2 - 10x - 5\sqrt{3} \)

Explanation:
1. A quadratic polynomial has the form \( ax^2 + bx + c \) with \( a \ne 0 \) and highest power 2.
2. \( \sqrt{3}x^2 - 10x - 5\sqrt{3} \) has highest power 2, while the others have degree 3 or 1.

Teacher's Note:
1. Look at the highest power of the variable.
2. Coefficients may be irrational numbers; only the power matters.

Answer: (d) \( x = 3 \)

Explanation:
1. At \( x = 3 \): \( 9 - 6 - 3 = 0 \), so 3 is a zero.
2. At \( x = -3 \): \( 9 + 6 - 3 = 12 \ne 0 \). At \( x = 2 \): \( 4 - 4 - 3 = -3 \ne 0 \). At \( x = 1 \): \( 1 - 2 - 3 = -4 \ne 0 \).

Teacher's Note:
1. A zero of a polynomial makes its value 0.
2. Substitute each option and check.

Answer: (a) 8

Explanation:
1. Substitute \( x = 2 \): \( 3(2)^2 - 2m + 4 = 0 \).
2. \( 12 - 2m + 4 = 0 \), so \( m = 8 \).

Teacher's Note:
1. If a is a zero, the polynomial has value 0 at x = a.
2. Check: 3x² - 8x + 4 at x = 2 gives 12 - 16 + 4 = 0.

Answer: (d) \( 5\sqrt{2} + 1 \)

Explanation:
1. \( \sqrt{2}(\sqrt{2})^2 + 3\sqrt{2} + 1 = 2\sqrt{2} + 3\sqrt{2} + 1 = 5\sqrt{2} + 1 \).

Teacher's Note:
1. (√2)² = 2, so √2 × 2 = 2√2.
2. Add the like surds: 2√2 + 3√2 = 5√2.

Answer: (b) \( -130 \)

Explanation:
1. \( (-4)^3 - 3(-4)^2 + 5(-4) + 2 = -64 - 48 - 20 + 2 = -130 \).

Teacher's Note:
1. Be careful with signs: (-4)³ = -64 but (-4)² = +16.
2. Add the four terms: -64 - 48 - 20 + 2.

Answer: (c) \( -\dfrac{7}{2} \)

Explanation:
1. \( 4(-2)^2 - 3m(-2) + 5 = 0 \) gives \( 16 + 6m + 5 = 0 \).
2. So \( 6m = -21 \) and \( m = -\dfrac{7}{2} \).

Teacher's Note:
1. Substitute the zero and solve the linear equation in m.
2. Mind the sign: -3m × (-2) = +6m.

Answer: (a) \( -\dfrac{29}{10} \)

Explanation:
1. \( 7 + 4(m+2)(-5) - 25 = 0 \), i.e. \( -20(m+2) - 18 = 0 \).
2. \( -20m - 58 = 0 \), so \( m = -\dfrac{29}{10} \).

Teacher's Note:
1. Substitute y = -5 carefully; y² = 25.
2. Collect constants: 7 - 25 = -18.

Answer: (c) \( m = -\dfrac{5}{8} \)

Explanation:
1. \( (2m + 1)\cdot 4 + 4 - 3 = 0 \) gives \( 8m + 5 = 0 \).
2. So \( m = -\dfrac{5}{8} \).

Teacher's Note:
1. Substitute x = 2 into the equation and solve for m.
2. 4(2m + 1) = 8m + 4.

Answer: (b) \( \dfrac{9}{4} \)

Explanation:
1. \( 3\cdot\dfrac{1}{4} + m - 3 = 0 \), so \( m = 3 - \dfrac{3}{4} = \dfrac{9}{4} \).

Teacher's Note:
1. 2m × (1/2) = m.
2. Check: 3x² + 4.5x - 3 at x = 1/2 gives 0.75 + 2.25 - 3 = 0.

Answer: (a) only \( x = \dfrac{3}{2} \)

Explanation:
1. At \( x = \frac{1}{2} \): \( \frac{1}{2} - \frac{5}{2} + 3 = 1 \ne 0 \).
2. At \( x = \frac{3}{2} \): \( \frac{9}{2} - \frac{15}{2} + 3 = 0 \).
3. So only \( \frac{3}{2} \) is a solution.

Teacher's Note:
1. The roots are 1 and 3/2, so 1/2 is not a root.
2. Substitute each given number.

Answer: (d) \( 3 \) and \( -6 \)

Explanation:
1. \( x^2 + 6x - 3x - 18 = x(x + 6) - 3(x + 6) = (x + 6)(x - 3) = 0 \).
2. So \( x = -6 \) or \( x = 3 \).

Teacher's Note:
1. Split the middle term: find two numbers with product -18 and sum 3.
2. The numbers are 6 and -3.

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

Explanation:
1. \( 2x^2 + 2ax - ax - a^2 = 2x(x + a) - a(x + a) = (x + a)(2x - a) = 0 \).
2. So \( x = -a \) or \( x = \dfrac{a}{2} \).

Teacher's Note:
1. Split ax as 2ax - ax.
2. Check the sum of roots: -a + a/2 = -a/2 = -(coefficient of x)/2.

Answer: (d) 0 and 3

Explanation:
1. \( x^2 - 3x = x(x - 3) = 0 \), so \( x = 0 \) or \( x = 3 \).

Teacher's Note:
1. Never divide by x: this loses the root x = 0.
2. Take everything to one side and factorise.

Answer: (c) \( 3 \) and \( -\dfrac{4}{3} \)

Explanation:
1. \( 3x^2 - 5x - 12 = 0 \), i.e. \( 3x^2 - 9x + 4x - 12 = 3x(x - 3) + 4(x - 3) = (x - 3)(3x + 4) = 0 \).
2. So \( x = 3 \) or \( x = -\dfrac{4}{3} \).

Teacher's Note:
1. Expand the bracket first.
2. Check: 3(9 - 4) = 15 = 5(3).

Answer: (d) \( 3 \) and \( -6 \)

Explanation:
1. Expanding: \( x^2 + x + x^2 + 5x + 6 = 42 \), so \( 2x^2 + 6x - 36 = 0 \), i.e. \( x^2 + 3x - 18 = 0 \).
2. \( (x + 6)(x - 3) = 0 \) gives \( x = -6 \) or \( x = 3 \).

Teacher's Note:
1. Expand both products carefully: (x+2)(x+3) = x² + 5x + 6.
2. Divide by 2 before factorising.

Answer: (a) \( 3 \) and \( \dfrac{1}{2} \)

Explanation:
1. Divide by 2: \( x^2 - \frac{7}{2}x + \frac{3}{2} = 0 \).
2. \( \left(x - \frac{7}{4}\right)^2 - \frac{49}{16} + \frac{3}{2} = 0 \), so \( \left(x - \frac{7}{4}\right)^2 = \frac{25}{16} \).
3. \( x - \frac{7}{4} = \pm\frac{5}{4} \) gives \( x = 3 \) or \( x = \frac{1}{2} \).

Teacher's Note:
1. Divide by the coefficient of x² first.
2. Add and subtract the square of half the coefficient of x.

Answer: (c) \( -\dfrac{\sqrt{3}}{2} \) and \( -\dfrac{\sqrt{3}}{2} \) (equal roots)

Explanation:
1. \( 4x^2 + 4\sqrt{3}x + 3 = (2x + \sqrt{3})^2 \).
2. So \( 2x + \sqrt{3} = 0 \) and \( x = -\dfrac{\sqrt{3}}{2} \) (twice).

Teacher's Note:
1. Check whether the expression is a perfect square.
2. D = 48 - 48 = 0 confirms equal roots.

Answer: (d) 4 and 3

Explanation:
1. \( a = 1, b = -7, c = 12 \); \( D = 49 - 48 = 1 \).
2. \( x = \dfrac{7 \pm 1}{2} = 4 \) or \( 3 \).

Teacher's Note:
1. Use x = (-b ± √D)/2a.
2. Check: sum 7 and product 12.

Answer: (c) \( 2 \) and \( -\dfrac{5}{3} \)

Explanation:
1. \( D = 1 + 120 = 121 \), so \( x = \dfrac{1 \pm 11}{2 \times 3} = \dfrac{12}{6} \) or \( \dfrac{-10}{6} \).
2. So \( x = 2 \) or \( x = -\dfrac{5}{3} \).
3. Check by factorising: \( (x - 2)(3x + 5) = 3x^2 - x - 10 \).

Teacher's Note:
1. The denominator in the formula is 2a = 6, not 2.
2. Check using the sum of roots: 2 - 5/3 = 1/3 = -b/a.

Answer: (c) \( \dfrac{\sqrt{5}}{3} \) and \( -\sqrt{5} \)

Explanation:
1. \( D = 20 + 60 = 80 \), \( \sqrt{D} = 4\sqrt{5} \).
2. \( x = \dfrac{-2\sqrt{5} \pm 4\sqrt{5}}{6} \), i.e. \( \dfrac{\sqrt{5}}{3} \) or \( -\sqrt{5} \).

Teacher's Note:
1. Simplify √80 = 4√5.
2. The product of roots is -5/3: (√5/3)(-√5) = -5/3.

Answer: (a) \( \dfrac{4b^2}{a^2} \) and \( -\dfrac{3a^2}{b^2} \)

Explanation:
1. \( D = (4b^4 - 3a^4)^2 + 48a^4b^4 = (4b^4 + 3a^4)^2 \).
2. \( x = \dfrac{(4b^4 - 3a^4) \pm (4b^4 + 3a^4)}{2a^2b^2} \), giving \( \dfrac{8b^4}{2a^2b^2} = \dfrac{4b^2}{a^2} \) and \( \dfrac{-6a^4}{2a^2b^2} = -\dfrac{3a^2}{b^2} \).

Teacher's Note:
1. Show that the discriminant is a perfect square.
2. Check: the product of roots is -12 = -12a²b²/(a²b²).

Answer: (b) \( -2 \) and \( -3 \)

Explanation:
1. \( (x + 2)(x + 3) = 0 \), so \( x = -2 \) or \( x = -3 \).

Teacher's Note:
1. Factorise or use the quadratic formula, and then check using the sum and product of roots.
2. Take care with signs of the coefficients.

Answer: (b) \( 11 \) and \( -3 \)

Explanation:
1. \( (x - 11)(x + 3) = 0 \), so \( x = 11 \) or \( x = -3 \).

Teacher's Note:
1. Factorise or use the quadratic formula, and then check using the sum and product of roots.
2. Take care with signs of the coefficients.

Answer: (a) \( a \) and \( -\dfrac{3a}{5} \)

Explanation:
1. \( 5x^2 - 5ax + 3ax - 3a^2 = 5x(x - a) + 3a(x - a) = (x - a)(5x + 3a) = 0 \).
2. So \( x = a \) or \( x = -\dfrac{3a}{5} \).

Teacher's Note:
1. Factorise or use the quadratic formula, and then check using the sum and product of roots.
2. Take care with signs of the coefficients.

Answer: (c) \( \dfrac{a}{3} \) and \( -a \)

Explanation:
1. \( 3x^2 + 3ax - ax - a^2 = 3x(x + a) - a(x + a) = (x + a)(3x - a) = 0 \).
2. So \( x = -a \) or \( x = \dfrac{a}{3} \).

Teacher's Note:
1. Factorise or use the quadratic formula, and then check using the sum and product of roots.
2. Take care with signs of the coefficients.