CBSE Class 10 Mathematics Quadratic Equations MCQs Set 27

Download CBSE MCQs for Class 10 Mathematics: Chapter 04 Quadratic Equations

Explore reliable objective questions for Chapter 04 Quadratic Equations tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Chapter-wise Objective Questions: Chapter 04 Quadratic Equations

View or download the dedicated Chapter 04 Quadratic Equations MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: The roots of \( 3a^2x^2 + 8abx + 4b^2 = 0 \) (a ≠ 0) are:
(a) \( \frac{2b}{3a},\ \frac{2b}{a} \)
(b) \( -\frac{2b}{3a},\ -\frac{2b}{a} \)
(c) \( -\frac{b}{3a},\ -\frac{4b}{a} \)
(d) \( -\frac{2b}{3a},\ \frac{2b}{a} \)
Show Answer & Explanation

Answer: (b) \( -\frac{2b}{3a},\ -\frac{2b}{a} \)

Explanation:
1. 3a²x² + 6abx + 2abx + 4b² = 0.
2. 3ax(ax + 2b) + 2b(ax + 2b) = 0, so (3ax + 2b)(ax + 2b) = 0.
3. x = −2b/(3a) or x = −2b/a.

Teacher's Note:
1. Split the middle term 8abx as 6abx + 2abx.
2. Equate each factor to zero.

Question 2: The roots of \( ax^2 + (4a^2 - 3b)x - 12ab = 0 \) (a ≠ 0) are:
(a) \( \frac{3b}{a},\ -4a \)
(b) \( \frac{3b}{a},\ 4a \)
(c) \( -\frac{3b}{a},\ -4a \)
(d) \( -\frac{3b}{a},\ 4a \)
Show Answer & Explanation

Answer: (a) \( \frac{3b}{a},\ -4a \)

Explanation:
1. Group: ax(x + 4a) − 3b(x + 4a) = 0.
2. (ax − 3b)(x + 4a) = 0.
3. x = 3b/a or x = −4a.

Teacher's Note:
1. Try the factors (ax − 3b) and (x + 4a).
2. Expand to confirm the middle term.

Question 3: Solving \( 2x^2 - 2\sqrt{6}\,x + 3 = 0 \) by completing the square gives:
(a) \( x = \sqrt{6}\ \text{(repeated)} \)
(b) \( x = \frac{\sqrt{6}}{2}\ \text{or}\ -\frac{\sqrt{6}}{2} \)
(c) \( x = \sqrt{3}\ \text{(repeated)} \)
(d) \( x = \frac{\sqrt{6}}{2}\ \text{(repeated)} \)
Show Answer & Explanation

Answer: (d) \( x = \frac{\sqrt{6}}{2}\ \text{(repeated)} \)

Explanation:
1. Divide by 2: x² − √6x + 3/2 = 0.
2. Add and subtract (√6/2)² = 3/2: (x − √6/2)² = 0.
3. x = √6/2, a repeated root.

Teacher's Note:
1. Check the discriminant: 24 − 24 = 0.
2. Write the left side as a perfect square.

Question 4: The roots of \( 9x^2 - 3(a^2 + b^2)x + a^2b^2 = 0 \) are:
(a) \( a^2,\ b^2 \)
(b) \( \frac{a}{3},\ \frac{b}{3} \)
(c) \( \frac{a^2}{3},\ \frac{b^2}{3} \)
(d) \( -\frac{a^2}{3},\ -\frac{b^2}{3} \)
Show Answer & Explanation

Answer: (c) \( \frac{a^2}{3},\ \frac{b^2}{3} \)

Explanation:
1. (3x − a²)(3x − b²) = 9x² − 3b²x − 3a²x + a²b².
2. This equals the given expression, so x = a²/3 or x = b²/3.

Teacher's Note:
1. Look for two factors whose constants multiply to a²b².
2. Verify by expanding.

Question 5: If −4 is a root of \( x^2 + px - 4 = 0 \) and the equation \( x^2 + px + q = 0 \) has equal roots, then p and q are:
(a) \( p = -3,\ q = \frac{9}{4} \)
(b) \( p = 3,\ q = \frac{9}{4} \)
(c) \( p = 3,\ q = \frac{4}{9} \)
(d) \( p = 4,\ q = 4 \)
Show Answer & Explanation

Answer: (b) \( p = 3,\ q = \frac{9}{4} \)

Explanation:
1. 16 − 4p − 4 = 0, so p = 3.
2. Discriminant of x² + 3x + q: 9 − 4q = 0.
3. q = 9/4.

Teacher's Note:
1. Substitute the known root first.
2. Equal roots mean the discriminant is zero.

Question 6: The value of k for which \( 4x^2 - 2x + k = 0 \) has one root equal to the reciprocal of the other is:
(a) 4
(b) 1/4
(c) −4
(d) 2
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Product of roots = k/4.
2. For reciprocal roots the product is 1, so k/4 = 1.
3. k = 4.

Teacher's Note:
1. Product of roots = c/a.
2. The product of a number and its reciprocal is 1.

Question 7: If α and β are roots of \( 2x^2 - 3x + 1 = 0 \), the value of \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \) is:
(a) 3/2
(b) 5
(c) 1/2
(d) 5/2
Show Answer & Explanation

Answer: (d) 5/2

Explanation:
1. α + β = 3/2, αβ = 1/2.
2. α² + β² = 9/4 − 1 = 5/4.
3. (5/4) ÷ (1/2) = 5/2.

Teacher's Note:
1. Write the sum as (α² + β²)/(αβ).
2. Use α² + β² = (α + β)² − 2αβ.

Question 8: If α and β are roots of \( ax^2 + bx + c = 0 \), the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) is:
(a) \( \frac{b}{c} \)
(b) \( -\frac{c}{b} \)
(c) \( -\frac{b}{c} \)
(d) \( \frac{c}{a} \)
Show Answer & Explanation

Answer: (c) \( -\frac{b}{c} \)

Explanation:
1. 1/α + 1/β = (α + β)/(αβ) = (−b/a)/(c/a) = −b/c.

Teacher's Note:
1. Combine the fractions.
2. Use the sum and product of roots.

Question 9: If α and β are roots of \( ax^2 + bx + c = 0 \), then \( \alpha^2\beta + \alpha\beta^2 \) equals:
(a) \( \frac{bc}{a^2} \)
(b) \( -\frac{bc}{a^2} \)
(c) \( -\frac{b^2}{ac} \)
(d) \( \frac{c^2}{a^2} \)
Show Answer & Explanation

Answer: (b) \( -\frac{bc}{a^2} \)

Explanation:
1. α²β + αβ² = αβ(α + β) = (c/a)(−b/a) = −bc/a².

Teacher's Note:
1. Factor out αβ.
2. Substitute the sum and product.

Question 10: If α and β are roots of \( ax^2 + bx + c = 0 \), the value of \( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} \) is:
(a) \( \frac{3abc - b^3}{a^2 c} \)
(b) \( \frac{b^3 - 3abc}{a^2 c} \)
(c) \( \frac{3abc + b^3}{a^2 c} \)
(d) \( \frac{3abc - b^3}{a c^2} \)
Show Answer & Explanation

Answer: (a) \( \frac{3abc - b^3}{a^2 c} \)

Explanation:
1. α³ + β³ = −b³/a³ + 3bc/a² = (3abc − b³)/a³.
2. Divide by αβ = c/a.
3. Result: (3abc − b³)/(a²c).

Teacher's Note:
1. Combine over αβ.
2. Use α³ + β³ = (α + β)³ − 3αβ(α + β).

Question 11: For which values of k does \( x^2 - (3k - 1)x + (2k^2 + 2k - 11) = 0 \) have real and equal roots?
(a) k = −5 or k = −9
(b) k = 5 only
(c) k = 9 only
(d) k = 5 or k = 9
Show Answer & Explanation

Answer: (d) k = 5 or k = 9

Explanation:
1. D = 9k² − 6k + 1 − 8k² − 8k + 44.
2. D = k² − 14k + 45 = (k − 5)(k − 9).
3. D = 0 gives k = 5 or 9.

Teacher's Note:
1. Set the discriminant to zero.
2. Factorise the resulting quadratic in k.

Question 12: The roots of \( x^2 - 4x + 1 = 0 \) are:
(a) \( 2 \pm \sqrt{5} \)
(b) \( 4 \pm \sqrt{3} \)
(c) \( 2 \pm \sqrt{3} \)
(d) \( -2 \pm \sqrt{3} \)
Show Answer & Explanation

Answer: (c) \( 2 \pm \sqrt{3} \)

Explanation:
1. x = (4 ± √(16 − 4))/2 = (4 ± 2√3)/2 = 2 ± √3.

Teacher's Note:
1. Use x = (−b ± √D)/(2a).
2. D = 16 − 4 = 12.

Question 13: The sum of the roots of \( \frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c} \) is zero. The product of the roots is:
(a) \( \frac{1}{2}(a^2 + b^2) \)
(b) \( -\frac{1}{2}(a^2 + b^2) \)
(c) \( -\frac{1}{2}(a + b)^2 \)
(d) \( -(a^2 + b^2) \)
Show Answer & Explanation

Answer: (b) \( -\frac{1}{2}(a^2 + b^2) \)

Explanation:
1. c(x + b) + c(x + a) = (x + a)(x + b).
2. x² + (a + b − 2c)x + ab − c(a + b) = 0.
3. Sum = −(a + b − 2c) = 0, so c = (a + b)/2.
4. Product = ab − (a + b)²/2 = −(a² + b²)/2.

Teacher's Note:
1. Clear the denominators and collect terms.
2. Use the zero sum to find c.

Question 14: The solutions of \( 7^{1+x} + 7^{1-x} = 50 \) are:
(a) x = 1 and x = −1
(b) x = 0 only
(c) x = 2 and x = −2
(d) x = 1 only
Show Answer & Explanation

Answer: (a) x = 1 and x = −1

Explanation:
1. 7·7^x + 7/7^x = 50.
2. 7t² − 50t + 7 = 0 with t = 7^x.
3. (7t − 1)(t − 7) = 0, so t = 1/7 or 7, giving x = −1 or 1.

Teacher's Note:
1. Put t = 7^x.
2. Multiply through by t to get a quadratic.

Question 15: The value(s) of k for which \( 2x^2 - kx + 1 = 0 \) has coincident roots is/are:
(a) \( k = 2\sqrt{2}\ \text{only} \)
(b) \( k = \pm 2 \)
(c) \( k = \pm 4 \)
(d) \( k = \pm 2\sqrt{2} \)
Show Answer & Explanation

Answer: (d) \( k = \pm 2\sqrt{2} \)

Explanation:
1. Coincident roots need k² − 8 = 0, so k = ±2√2.

Teacher's Note:
1. Set D = b² − 4ac = 0.
2. Remember both signs.

Question 16: The roots of \( \frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4} \) are:
(a) \( -2 \pm 2\sqrt{3} \)
(b) \( 2 \pm \sqrt{3} \)
(c) \( 2 \pm 2\sqrt{3} \)
(d) \( 4 \pm 2\sqrt{3} \)
Show Answer & Explanation

Answer: (c) \( 2 \pm 2\sqrt{3} \)

Explanation:
1. (x² + 6x + 8) + 2(x² + 5x + 4) = 4(x² + 3x + 2).
2. 3x² + 16x + 16 = 4x² + 12x + 8.
3. x² − 4x − 8 = 0, so x = 2 ± 2√3.

Teacher's Note:
1. Multiply by (x+1)(x+2)(x+4).
2. Collect terms to get a quadratic.

Question 17: If the roots of \( p(q-r)x^2 + q(r-p)x + r(p-q) = 0 \) are equal, then:
(a) \( \frac{2}{q} = \frac{1}{p} - \frac{1}{r} \)
(b) \( \frac{2}{q} = \frac{1}{p} + \frac{1}{r} \)
(c) \( q = p + r \)
(d) \( \frac{1}{q} = \frac{1}{p} + \frac{1}{r} \)
Show Answer & Explanation

Answer: (b) \( \frac{2}{q} = \frac{1}{p} + \frac{1}{r} \)

Explanation:
1. Sum of coefficients = pq − pr + qr − pq + pr − qr = 0, so x = 1 is a root.
2. Equal roots: product = 1, so r(p − q) = p(q − r).
3. rp − rq = pq − pr gives 2pr = q(p + r), i.e. 2/q = 1/p + 1/r.

Teacher's Note:
1. x = 1 is always a root of this equation.
2. Equal roots means the other root is also 1.

Question 18: The real solutions of \( 4x^4 - 4x^3 - 7x^2 - 4x + 4 = 0 \) are:
(a) \( x = \frac{1}{2},\ 2 \)
(b) \( x = -\frac{1}{2},\ -2 \)
(c) \( x = 1,\ 4 \)
(d) \( x = \frac{1}{2},\ -2 \)
Show Answer & Explanation

Answer: (a) \( x = \frac{1}{2},\ 2 \)

Explanation:
1. x² + 1/x² = y² − 2.
2. 4(y² − 2) − 4y − 7 = 0 gives 4y² − 4y − 15 = 0, so y = 5/2 or −3/2.
3. x + 1/x = 5/2 gives 2x² − 5x + 2 = 0, so x = 2 or 1/2.
4. x + 1/x = −3/2 gives a negative discriminant, so no real root.

Teacher's Note:
1. Divide by x² and put y = x + 1/x.
2. x + 1/x = −3/2 has no real solution.

Question 19: The equation \( 8\sec^2\theta - 6\sec\theta + 1 = 0 \) has how many real solutions for θ?
(a) 1
(b) 2
(c) 4
(d) 0
Show Answer & Explanation

Answer: (d) 0

Explanation:
1. (4 sec θ − 1)(2 sec θ − 1) = 0.
2. sec θ = 1/4 or 1/2.
3. Both have absolute value below 1, impossible for sec θ.

Teacher's Note:
1. Solve for sec θ first.
2. Recall the range of sec θ.

Question 20: If the roots of \( ax^2 + 2(a+b)x + (a + 2b + c) = 0 \) are imaginary, then the roots of \( ax^2 + 2bx + c = 0 \) are:
(a) real and equal
(b) real and distinct
(c) imaginary (non-real)
(d) cannot be determined
Show Answer & Explanation

Answer: (c) imaginary (non-real)

Explanation:
1. D₁ = 4(a + b)² − 4a(a + 2b + c) = 4(b² − ac).
2. D₁ < 0 gives b² − ac < 0.
3. D₂ = 4b² − 4ac = D₁ < 0, so imaginary roots.

Teacher's Note:
1. Both discriminants are the same expression.
2. Compare b² with ac.

Question 21: The condition that one root of \( ax^2 + bx + c = 0 \) is n times the other is:
(a) \( nb^2 = (n-1)^2 ac \)
(b) \( nb^2 = (n+1)^2 ac \)
(c) \( b^2 = n(n+1)ac \)
(d) \( n b = (n+1)^2 ac \)
Show Answer & Explanation

Answer: (b) \( nb^2 = (n+1)^2 ac \)

Explanation:
1. (n+1)α = −b/a and nα² = c/a.
2. Square the first: (n+1)²α² = b²/a².
3. Divide by the second: (n+1)²/n = b²/(ac), so nb² = (n+1)²ac.

Teacher's Note:
1. Let the roots be α and nα.
2. Eliminate α using the sum and product.

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