Quadratic Equations: MCQs with Answers

Official CBSE answer · Solution by Team TutorVertex

x² + 3x − 10 = (x + 5)(x − 2)

x + 5 = 0 or x − 2 = 0

Roots: −5 and 2.

Correct answer: A — 2, −5

Official CBSE answer · Solution by Team TutorVertex

x² − 25 = (x − 5)(x + 5)

x − 5 = 0 or x + 5 = 0

Roots: 5 and −5.

Correct answer: B — −5, 5

Official CBSE answer · Solution by Team TutorVertex

a = 3, b = −2√6, c = 2

b² − 4ac = (−2√6)² − 4(3)(2)

= 24 − 24 = 0

Hence the roots are real and equal.

Correct answer: C — Real and equal

Official CBSE answer · Solution by Team TutorVertex

a = 1, b = 1, c = 1

b² − 4ac = 1 − 4 = −3

Since −3 < 0, there are no real roots.

Correct answer: D — Not real

Official CBSE answer · Solution by Team TutorVertex

b² − 4ac = (−5)² − 4(4)(4)

= 25 − 64 = −39

Since −39 < 0, the roots are not real.

Correct answer: C — Not real

Official CBSE answer · Solution by Team TutorVertex

a = 1, b = −4, c = 3

b² − 4ac = (−4)² − 4(1)(3)

= 16 − 12 = 4

Correct answer: C — 4

Official CBSE answer · Solution by Team TutorVertex

b² − 4ac = (−6)² − 4(9)(−2)

= 36 + 72 = 108

Since 108 > 0, there are two distinct real roots.

Correct answer: C — Two distinct real roots

Official CBSE answer · Solution by Team TutorVertex

b² − 4ac = (−4√3)² − 4(3)(4)

= 48 − 48 = 0

The roots are real and equal.

Correct answer: B — Real and equal

Official CBSE answer · Solution by Team TutorVertex

b² − 4ac = 1² − 4(1)(−1) = 5

5 > 0 and √5 is irrational.

Roots = (−1 ± √5)/2

Hence the roots are irrational and distinct.

Correct answer: A — Irrational and distinct

Official CBSE answer · Solution by Team TutorVertex

b² − 4ac = 16

(−2)² − 4(3)c = 16

4 − 12c = 16

−12c = 12 ⇒ c = −1

Correct answer: C — −1

Official CBSE answer · Solution by Team TutorVertex

Given: y = 1

p(1)² + p(1) + 3 = 0

2p + 3 = 0

p = $\frac{\text{−3}}{\text{2}}$

Correct answer: C — $\frac{\text{−3}}{\text{2}}$

Official CBSE answer · Solution by Team TutorVertex

Equal roots: b² − 4ac = 0

(−9k)² − 4(5)(5) = 0

81k² = 100

k² = $\frac{\text{100}}{\text{81}}$

k = ±$\frac{\text{10}}{\text{9}}$

Correct answer: D — ±$\frac{\text{10}}{\text{9}}$

Official CBSE answer · Solution by Team TutorVertex

Equal roots: b² − 4ac = 0

4² − 4(1)k = 0

16 − 4k = 0

k = 4

Correct answer: C — k = 4

Official CBSE answer · Solution by Team TutorVertex

Substitute x = 3:

9 − 6k − 6 = 0

3 − 6k = 0

k = $\frac{\text{1}}{\text{2}}$

Correct answer: B — $\frac{\text{1}}{\text{2}}$

Official CBSE answer · Solution by Team TutorVertex

Equal roots: b² − 4ac = 0

(−k)² − 4(2)k = 0

k² − 8k = 0

k(k − 8) = 0

k = 0 or k = 8

Correct answer: D — 0, 8

Official CBSE answer · Solution by Team TutorVertex

For k ≠ 0: b² − 4ac = 25 − 4k

No real roots requires 25 − 4k < 0

k > $\frac{\text{25}}{\text{4}}$

Only 7 among the listed values satisfies this.

At k = 0 the equation is linear and has the real solution x = $\frac{\text{1}}{\text{5}}$.

Correct answer: D — 7

Official CBSE answer · Solution by Team TutorVertex

Sum of roots = (2 + √3) + (2 − √3) = 4

Product = (2 + √3)(2 − √3) = 4 − 3 = 1

Equation: x² − (sum)x + product = 0

x² − 4x + 1 = 0

Correct answer: A — x² − 4x + 1 = 0

Official CBSE answer · Solution by Team TutorVertex

Sum of roots = 6

Product = 3² − (√2)² = 9 − 2 = 7

Equation: x² − 6x + 7 = 0

Correct answer: A — x² − 6x + 7 = 0

Official CBSE answer · Solution by Team TutorVertex

First root = 2

Second root = 0 − 2 = −2

Equation: (x − 2)(x + 2) = 0

x² − 4 = 0

Correct answer: D — x² − 4 = 0

Official CBSE answer · Solution by Team TutorVertex

Positive factors of prime p: 1 and p

Sum = p + 1; product = p

Equation: (x − 1)(x − p) = 0

x² − (p + 1)x + p = 0

Correct answer: B — x² − (p + 1)x + p = 0

Official CBSE solution · typesetting normalised

(x + 2)(x + 1) = x² + 2x + 3

x² + 3x + 2 = x² + 2x + 3

x − 1 = 0

It is not a quadratic equation.

Correct answer: D — (x + 2)(x + 1) = x² + 2x + 3

Official CBSE solution · typesetting normalised

On simplification,

(C) 3x + 1 = 0

NOT a Quadratic Equation.

Correct answer: C — (x + 2)(x − 1) = x² − 2x − 3