CBSE Maths › Class 10 PYQs › Quadratic Equations

Quadratic Equations: 1 mark Questions (CBSE Class 10)

73 different 1 mark questions on Quadratic Equations from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (73)2 marks (38)3 marks (23)4 marks (13)5 marks (62)

Q21 markMCQQuadratic Equations CBSE 2026 · Basic 430/4/1

The value of k for which the equation $2 k x^{2} - 6 x + 3 = 0$ has real and equal roots, is :

(A) $\frac{3}{2}$
(B) $\frac{1}{2}$
(C) $- \frac{3}{2}$
(D)2

Show answer & solution

Answer: (A)

\frac{3}{2}

For equal roots, $D = b^{2} - 4 a c = 0$ .
$36 - 4 (2 k) (3) = 0 \Rightarrow 36 - 24 k = 0$ .
$k = \frac{3}{2}$ .

Also asked in: 2026 Basic 430/4/3

Q181 markMCQQuadratic Equations CBSE 2026 · Basic 430/4/2

The value of k for which the equation $5 x^{2} - 2 x + k = 0$ has equal real roots, is :

(A)5
(B) $- \frac{1}{5}$
(C) $\frac{1}{5}$
(D)0

Show answer & solution

Answer: (C)

\frac{1}{5}

For equal roots, $b^{2} - 4 a c = 0$ .
$4 - 20 k = 0$ .
$k = \frac{1}{5}$ .

Q21 markMCQQuadratic Equations CBSE 2026 · Basic 430/5/1

The roots of the quadratic equation $x^{2} + 9 = 0$ are

(A)real and equal
(B)not real
(C)real and negative of each other
(D)rational numbers

Show answer & solution

Answer: (B) not real

Here $a = 1$ , $b = 0$ , $c = 9$ .
$D = b^{2} - 4 a c = 0 - 36 = - 36 < 0$ .
So the roots are not real.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q11 markMCQQuadratic Equations CBSE 2026 · Standard 30/3/1

The roots of the quadratic equation $(x - 1)^{2} = 16$ are :

(A)5, 3
(B)4, – 4
(C)5, – 3
(D)– 5, 3

Show answer & solution

Answer: (C) 5, – 3

$(x - 1)^{2} = 16$ gives $x - 1 = \pm 4$
$x = 1 + 4 = 5$ or $x = 1 - 4 = - 3$
Roots: 5 and $- 3$

Also asked in: 2026 Standard 30/3/3

Q101 markMCQQuadratic Equations CBSE 2026 · Standard 30/3/1

If the roots of the quadratic equation $3 x^{2} - k x + 23 = 0$ are real and equal, then the value(s) of $k$ is/are :

(A) $\pm 24$
(B)0
(C)4
(D)– 5

Show answer & solution

Answer: (A)

\pm 24

Equal roots: $D = b^{2} - 4 a c = 0$
$k^{2} - 4 \times 3 \times 23 = 0$ , so $k^{2} = 24$
$k = \pm 24$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q71 markMCQQuadratic Equations CBSE 2026 · Standard 30/3/2

The roots of the quadratic equation $4 x^{2} - (a - 1)^{2} = 0$ are :

(A) $a - 1, a + 1$
(B) $\frac{a - 1}{2}, \frac{- a + 1}{2}$
(C) $\frac{a - 1}{2}, \frac{- a - 1}{2}$
(D) $\pm (a - 1)$

Show answer & solution

Answer: (B)

\frac{a - 1}{2}, \frac{- a + 1}{2}

$4 x^{2} = (a - 1)^{2}$ , so $(2 x)^{2} = (a - 1)^{2}$
$2 x = \pm (a - 1)$
$x = \frac{a - 1}{2}$ or $x = \frac{- ( a - 1 )}{2} = \frac{- a + 1}{2}$

Q11 markMCQQuadratic Equations CBSE 2026 · Standard 30/4/1

If the quadratic equation $9 x^{2} + 8 k x + 16 = 0$ has real and equal roots, then the value of k is

(A)3
(B) $- 3$
(C) $- 4$
(D) $\frac{3}{2}$

Show answer & solution

Answer: (A) 3

For equal roots, $D = b^{2} - 4 a c = 0$ .
$(8 k)^{2} - 4 \times 9 \times 16 = 0 \Rightarrow 64 k^{2} = 576$
$k^{2} = 9 \Rightarrow k = \pm 3$
Both 3 and $- 3$ are valid; option (A) 3 (option (B) $- 3$ is also correct).

Also asked in: 2026 Standard 30/4/3

Q111 markMCQQuadratic Equations CBSE 2026 · Standard 30/4/2

If roots of the quadratic equation $x^{2} - k 3 x + 2 = 0$ are real and equal, then value of k is

(A) $- 2$
(B) $\frac{8}{3}$
(C)1
(D)2

Show answer & solution

Answer: (B)

\frac{8}{3}

Equal roots: $(k 3)^{2} - 4 \times 1 \times 2 = 0$
$3 k^{2} = 8 \Rightarrow k = \pm \frac{8}{3}$
Of the options, only $\frac{8}{3}$ fits.

Q11 markMCQQuadratic Equations CBSE 2026 · Standard 30/5/1

The value of k for which the equation $k x^{2} - 6 x - 4 = 0$ has real and equal roots, is

(A) $\frac{9}{4}$
(B) $- 4$
(C) $- \frac{9}{4}$
(D) $- 2$

Show answer & solution

Answer: (C)

- \frac{9}{4}

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$(- 6)^{2} - 4 (k) (- 4) = 0$
$36 + 16 k = 0$
$k = - \frac{9}{4}$

Also asked in: 2026 Standard 30/5/2

Q91 markMCQQuadratic Equations CBSE 2026 · Standard 30/5/1

The value of p for which roots of the quadratic equation $x^{2} - p x + 6 = 0$ are rational, is

(A)1
(B) $- 5$
(C)25
(D) $5$

Show answer & solution

Answer: (B)

- 5

Roots are rational when $D = p^{2} - 24$ is a perfect square (and p is rational).
$p = 1$ : $D = - 23$ (no real roots).
$p = - 5$ : $D = 25 - 24 = 1 = 1^{2}$ , a perfect square; $x^{2} + 5 x + 6 = 0$ gives $x = - 2, - 3$ , which are rational.
$p = 25$ : $D = 601$ , not a perfect square. $p = 5$ : $D = - 19$ .
So $p = - 5$ .

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q141 markMCQQuadratic Equations CBSE 2026 · Standard 30/5/3

The value of m for which the quadratic equation $3 x^{2} - 7 x + m = 0$ has real and equal roots, is

(A)7
(B) $\frac{49}{12}$
(C) $\frac{49}{3}$
(D)4

Show answer & solution

Answer: (B)

\frac{49}{12}

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$49 - 12 m = 0$
$m = \frac{49}{12}$

Q31 markMCQQuadratic Equations CBSE 2025 · Basic 430/1/1

The discriminant of the quadratic equation $x^{2} - 3 x - 2 = 0$ is :

(A)1
(B)17
(C) $17$
(D) $- 17$

Show answer & solution

Answer: (B) 17

Here $a = 1$ , $b = - 3$ , $c = - 2$ .
$D = b^{2} - 4 a c = 9 + 8 = 17$ .

Q41 markMCQQuadratic Equations CBSE 2025 · Basic 430/1/1

The equation $x + \frac{1}{x} = 3$ $(x \neq = 0)$ is expressed as a quadratic equation in the form of $a x^{2} + b x + c = 0$ . The value of $a - b + c$ is :

(A)5
(B)2
(C)1
(D) $- 1$

Show answer & solution

Answer: (A) 5

Multiply by $x$ : $x^{2} + 1 = 3 x$ , i.e. $x^{2} - 3 x + 1 = 0$ .
So $a = 1$ , $b = - 3$ , $c = 1$ .
$a - b + c = 1 + 3 + 1 = 5$ .

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q151 markMCQQuadratic Equations CBSE 2025 · Basic 430/1/2

The discriminant of the quadratic equation $- x^{2} - 5 x + 6 = 0$ is :

(A)1
(B) $- 1$
(C)49
(D)7

Show answer & solution

Answer: (C) 49

Here $a = - 1$ , $b = - 5$ , $c = 6$ .
$D = b^{2} - 4 a c = 25 - 4 (- 1) (6) = 25 + 24 = 49$ .

Q81 markMCQQuadratic Equations CBSE 2025 · Basic 430/1/3

The discriminant of the quadratic equation $2 x^{2} - 3 x - 5 = 0$ is :

(A) $- 31$
(B)49
(C)7
(D) $- 31$

Show answer & solution

Answer: (B) 49

Here $a = 2$ , $b = - 3$ , $c = - 5$ .
$D = b^{2} - 4 a c = 9 + 40 = 49$ .

Q41 markMCQQuadratic Equations CBSE 2025 · Basic 430/2/1

The value of ‘a’ for which $a x^{2} + 3 x + 1 = 0$ has real and equal roots is :

(A) $\frac{4}{9}$
(B) $\frac{9}{4}$
(C) $\frac{3}{2}$
(D) $\frac{2}{3}$

Show answer & solution

Answer: (B)

\frac{9}{4}

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$3^{2} - 4 (a) (1) = 0 \Rightarrow 9 - 4 a = 0$ .
So $a = \frac{9}{4}$ .

Q51 markMCQQuadratic Equations CBSE 2025 · Basic 430/2/1

Which of the following equations is a quadratic equation ?

(A) $x^{2} = (x + 1)^{2}$
(B) $(x - 1) (x + 2) = 2 x + 1$
(C) $(x + 2)^{3} = 2 x (x^{2} - 1)$
(D) $x = x^{2}$

Show answer & solution

Answer: (B)

(x - 1) (x + 2) = 2 x + 1

(A) $x^{2} = x^{2} + 2 x + 1$ gives $2 x + 1 = 0$ , which is linear.
(B) $x^{2} + x - 2 = 2 x + 1$ gives $x^{2} - x - 3 = 0$ , which is quadratic.
(C) $x^{3} + 6 x^{2} + 12 x + 8 = 2 x^{3} - 2 x$ gives $x^{3} - 6 x^{2} - 14 x - 8 = 0$ , which is cubic.
(D) $x = x^{2}$ is not a polynomial equation.
So (B) is the quadratic equation.

Also asked in: 2025 Basic 430/2/2, 2025 Basic 430/2/3

Q161 markMCQQuadratic Equations CBSE 2025 · Basic 430/2/2

The value of ‘a’ for which the equation $x^{2} + 3 x + a = 0$ has real and equal roots is :

(A) $\frac{4}{9}$
(B) $\frac{9}{4}$
(C) $\frac{3}{2}$
(D) $\frac{2}{3}$

Show answer & solution

Answer: (B)

\frac{9}{4}

For real and equal roots, $b^{2} - 4 a c = 0$ .
$9 - 4 a = 0 \Rightarrow a = \frac{9}{4}$ .

Q91 markMCQQuadratic Equations CBSE 2025 · Basic 430/2/3

One of the values of ‘p’ for which $p x^{2} + 4 x + p = 0$ has real and equal roots is :

(A) $4$
(B) $- 4$
(C) $2$
(D) $0$

Show answer & solution

Answer: (C) 2

For real and equal roots, $b^{2} - 4 a c = 0$ : $16 - 4 p \cdot p = 0$ .
$p^{2} = 4 \Rightarrow p = \pm 2$ (p = 0 is not allowed for a quadratic).
From the options, $p = 2$ .

Q51 markMCQQuadratic Equations CBSE 2025 · Basic 430/3/1

If $x = x$ , $(x \neq = 0)$ is expressed as a quadratic equation in the form $a x^{2} + b x + c = 0$ , then the value of $a + b + c$ is :

(A)0
(B)1
(C)2
(D)3

Show answer & solution

Answer: (A) 0

Squaring both sides: $x^{2} = x$ , i.e. $x^{2} - x = 0$ .
So $a = 1$ , $b = - 1$ , $c = 0$ .
$a + b + c = 1 - 1 + 0 = 0$ .

Q201 markAssertion–ReasonQuadratic Equations CBSE 2025 · Basic 430/3/1

Assertion (A) : Every quadratic equation has two real roots.
Reason (R) : A quadratic polynomial can have at most two zeroes.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

Show answer & solution

Answer: (D) Assertion (A) is false, but Reason (R) is true.

A: $x^{2} + 1 = 0$ has no real roots, so A is false.
R: a polynomial of degree 2 has at most 2 zeroes, so R is true.

Also asked in: 2025 Basic 430/3/2, 2025 Basic 430/3/3

Q171 markMCQQuadratic Equations CBSE 2025 · Basic 430/3/2

If $(x + 1)^{2} = x^{2} + 2 x$ is expressed as a quadratic equation in the form of $a x^{2} + b x + c = 0$ , then the value of $a - b + c$ is :

(A) $- 1$
(B)0
(C)1
(D)2

Show answer & solution

Answer: (C) 1

$(x + 1)^{2} = x + 2 x + 1$ .
So $x + 2 x + 1 = x^{2} + 2 x$ , i.e. $x^{2} - x - 1 = 0$ .
$a = 1$ , $b = - 1$ , $c = - 1$ .
$a - b + c = 1 + 1 - 1 = 1$ .

Q101 markMCQQuadratic Equations CBSE 2025 · Basic 430/3/3

If $(x + 1)^{3} = x^{3} + 1$ is expressed as a quadratic equation in the form $p x^{2} + q x + r = 0$ , then the value of $p - q + r$ is :

(A)0
(B)1
(C)3
(D)6

Show answer & solution

Answer: (A) 0

$(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1$ .
So $x^{3} + 3 x^{2} + 3 x + 1 = x^{3} + 1$ , i.e. $3 x^{2} + 3 x = 0$ .
$p = 3$ , $q = 3$ , $r = 0$ , so $p - q + r = 3 - 3 + 0 = 0$ .

Q21 markMCQQuadratic Equations CBSE 2025 · Basic 430/5/1

The value of k for which the roots of the quadratic equation $6 x^{2} + 4 k x + k = 0$ are real and equal, is

(A)0
(B) $\frac{3}{4}$
(C) $\frac{- 3}{2}$
(D) $\frac{2}{3}$

Show answer & solution

Answer: (A) 0

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$(4 k)^{2} - 4 \times 6 \times k = 16 k^{2} - 24 k = 8 k (2 k - 3) = 0$ .
So $k = 0$ or $k = \frac{3}{2}$ .
Of the options only $k = 0$ fits (the equation becomes $6 x^{2} = 0$ with equal roots 0, 0).

Also asked in: 2025 Basic 430/5/3

Q71 markMCQQuadratic Equations CBSE 2025 · Basic 430/5/1

The roots of the equation $x^{2} - 8 = 0$ are

(A)rational and distinct
(B)irrational and distinct
(C)real and equal
(D)not real

Show answer & solution

Answer: (B) irrational and distinct

$x^{2} = 8$ , so $x = \pm 22$ .
The two roots are different and irrational.

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q121 markMCQQuadratic Equations CBSE 2025 · Basic 430/5/2

The value of k for which roots of quadratic equation k $x$ ( $x$ – 2) + 6 = 0 are real and equal, is

(A)0 only
(B)0, 6
(C)6 only
(D)–6 only

Show answer & solution

Answer: (C) 6 only

The equation is $k x^{2} - 2 k x + 6 = 0$ .
For real and equal roots, $D = (- 2 k)^{2} - 4 \times k \times 6 = 4 k^{2} - 24 k = 4 k (k - 6) = 0$ , so $k = 0$ or $k = 6$ .
For $k = 0$ the equation is not quadratic (it becomes $6 = 0$ ), so $k = 6$ only.

Q31 markMCQQuadratic Equations CBSE 2025 · Standard 30/2/1

The quadratic equation whose roots are 7 and $\frac{1}{7}$ is :

(A) $7 x^{2} - 50 x + 7 = 0$
(B) $7 x^{2} - 50 x + 1 = 0$
(C) $7 x^{2} + 50 x - 7 = 0$
(D) $7 x^{2} + 50 x - 1 = 0$

Show answer & solution

Answer: (A)

7 x^{2} - 50 x + 7 = 0

Sum of roots $= 7 + \frac{1}{7} = \frac{50}{7}$ , product $= 7 \times \frac{1}{7} = 1$
Equation: $x^{2} - \frac{50}{7} x + 1 = 0$
Multiply by 7: $7 x^{2} - 50 x + 7 = 0$

Also asked in: 2025 Standard 30/2/3

Q11 markMCQQuadratic Equations CBSE 2025 · Standard 30/2/2

The quadratic equation whose sum and product of roots are ‘ $a$ ’ and ‘ $\frac{1}{a}$ ’ respectively is :

(A) $a x^{2} - a x + 1 = 0$
(B) $a x^{2} - a^{2} x + 1 = 0$
(C) $a x^{2} + a x + 1 = 0$
(D) $a x^{2} + a^{2} x - 1 = 0$

Show answer & solution

Answer: (B)

a x^{2} - a^{2} x + 1 = 0

Equation: $x^{2} - (sum) x + product = 0$
$x^{2} - a x + \frac{1}{a} = 0$
Multiply by a: $a x^{2} - a^{2} x + 1 = 0$

Q51 markMCQQuadratic Equations CBSE 2025 · Standard 30/3/1

If $\frac{x}{12} - \frac{3}{x} = 0$ , then the values of $x$ are :

(A) $\pm 6$
(B) $\pm 4$
(C) $\pm 12$
(D) $\pm 3$

Show answer & solution

Answer: (A)

\pm 6

$\frac{x}{12} = \frac{3}{x}$ gives $x^{2} = 36$ .
$x = \pm 6$ .

Also asked in: 2025 Standard 30/3/3

Q141 markMCQQuadratic Equations CBSE 2025 · Standard 30/3/2

The discriminant of the quadratic equation $b x^{2} + a x + c = 0$ ; $b \neq = 0$ is given by :

(A) $b^{2} - 4 a c$
(B) $b^{2} - 4 a c$
(C) $a^{2} - 4 b c$
(D) $a^{2} - 4 b c$

Show answer & solution

Answer: (D)

a^{2} - 4 b c

Here the coefficient of $x^{2}$ is $b$ , of $x$ is $a$ , and the constant term is $c$ .
Discriminant $= (coefficient of x)^{2} - 4 (coefficient of x^{2}) (constant) = a^{2} - 4 b c$ .

Q31 markMCQQuadratic Equations CBSE 2025 · Standard 30/4/1

The value of ' $a$ ' for which $a x^{2} + x + a = 0$ has equal and positive roots is :

(A) $2$
(B) $- 2$
(C) $\frac{1}{2}$
(D) $- \frac{1}{2}$

Show answer & solution

Answer: (D)

- \frac{1}{2}

Equal roots: $D = 1 - 4 a^{2} = 0$ , so $a = \pm \frac{1}{2}$
Each root $= - \frac{1}{2 a}$
For a positive root, $a < 0$ , so $a = - \frac{1}{2}$ (root $= 1$ ).

Also asked in: 2025 Standard 30/4/2, 2025 Standard 30/4/3

Q41 markMCQQuadratic Equations CBSE 2025 · Standard 30/5/1

Which of the following equations is a quadratic equation ?

(A) $x^{2} + 1 = (x - 1)^{2}$
(B) $(x + x)^{2} = 2 x x$
(C) $x^{3} + 3 x^{2} = (x + 1)^{3}$
(D) $(x + 1) (x - 1) = (x + 1)^{2}$

Show answer & solution

Answer: (B)

(x + x)^{2} = 2 x x

(A) $x^{2} + 1 = x^{2} - 2 x + 1 \Rightarrow 2 x = 0$ : linear.
(B) $x^{2} + 2 x x + x = 2 x x \Rightarrow x^{2} + x = 0$ : quadratic.
(C) $x^{3} + 3 x^{2} = x^{3} + 3 x^{2} + 3 x + 1 \Rightarrow 3 x + 1 = 0$ : linear.
(D) $x^{2} - 1 = x^{2} + 2 x + 1 \Rightarrow 2 x + 2 = 0$ : linear.
So (B).

Q51 markMCQQuadratic Equations CBSE 2025 · Standard 30/5/1

If $x^{2} + b x + b = 0$ has two real and distinct roots, then the value of $b$ can be

(A)0
(B)4
(C)3
(D) $- 3$

Show answer & solution

Answer: (D)

- 3

Real and distinct roots: $D = b^{2} - 4 b > 0 \Rightarrow b (b - 4) > 0$
So $b < 0$ or $b > 4$ .
Among the options only $b = - 3$ works.

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q81 markMCQQuadratic Equations CBSE 2025 · Standard 30/5/2

Which of the following equations is a quadratic equation ?

(A) $x^{3} = (x - 1)^{3} + 3 x^{2}$
(B) $x^{3} = (x + 1)^{3}$
(C) $x^{2} = x$
(D) $x^{2} + 1 = \frac{1}{x}$

Show answer & solution

Answer: (B)

x^{3} = (x + 1)^{3}

(A) $x^{3} = x^{3} - 3 x^{2} + 3 x - 1 + 3 x^{2} \Rightarrow 3 x - 1 = 0$ : linear.
(B) $x^{3} = x^{3} + 3 x^{2} + 3 x + 1 \Rightarrow 3 x^{2} + 3 x + 1 = 0$ : quadratic.
(C) $x^{2} = x$ is not a polynomial equation of degree 2.
(D) $x^{2} + 1 = \frac{1}{x} \Rightarrow x^{3} + x - 1 = 0$ : cubic.
So (B).

Q131 markMCQQuadratic Equations CBSE 2025 · Standard 30/5/3

Which of the following equation is a quadratic equation ?

(A) $(x + \frac{1}{x})^{2} = 2$
(B) $(x - x)^{2} + 2 x x = 0$
(C) $(x + 1)^{3} = (1 - x)^{3}$
(D) $(x + 1)^{2} = x^{2}$

Show answer & solution

Answer: (B)

(x - x)^{2} + 2 x x = 0

(A) $x^{2} + 2 + \frac{1}{x ^{2}} = 2 \Rightarrow x^{4} + 1 = 0$ : degree 4.
(B) $x^{2} - 2 x x + x + 2 x x = 0 \Rightarrow x^{2} + x = 0$ : quadratic.
(C) $x^{3} + 3 x^{2} + 3 x + 1 = 1 - 3 x + 3 x^{2} - x^{3} \Rightarrow 2 x^{3} + 6 x = 0$ : cubic.
(D) $x + 2 x + 1 = x^{2}$ contains $x$ : not a quadratic equation.
So (B).

Q31 markMCQQuadratic Equations CBSE 2025 · Standard 30/6/1

Which of the following quadratic equations has real and equal roots ?

(A) $(x + 1)^{2} = 2 x + 1$
(B) $x^{2} + x = 0$
(C) $x^{2} - 4 = 0$
(D) $x^{2} + x + 1 = 0$

Show answer & solution

Answer: (A)

(x + 1)^{2} = 2 x + 1

(A) $x^{2} + 2 x + 1 = 2 x + 1 \Rightarrow x^{2} = 0$ ; $D = 0$ , so roots are real and equal ( $x = 0, 0$ ).
(B) $x^{2} + x = 0$ : $D = 1 > 0$ , distinct roots.
(C) $x^{2} - 4 = 0$ : $D = 16 > 0$ , distinct roots.
(D) $x^{2} + x + 1 = 0$ : $D = 1 - 4 = - 3 < 0$ , no real roots.

Q51 markMCQQuadratic Equations CBSE 2025 · Standard 30/6/2

Which of the following quadratic equations has real and distinct roots ?

(A) $x^{2} + 2 x = 0$
(B) $x^{2} + x + 1 = 0$
(C) $(x - 1)^{2} = 1 - 2 x$
(D) $2 x^{2} + x + 1 = 0$

Show answer & solution

Answer: (A)

x^{2} + 2 x = 0

(A) $x^{2} + 2 x = 0$ : $D = 4 - 0 = 4 > 0$ , real and distinct roots ( $0, - 2$ ).
(B) $x^{2} + x + 1 = 0$ : $D = 1 - 4 = - 3 < 0$ , no real roots.
(C) $(x - 1)^{2} = 1 - 2 x \Rightarrow x^{2} - 2 x + 1 = 1 - 2 x \Rightarrow x^{2} = 0$ : equal roots.
(D) $2 x^{2} + x + 1 = 0$ : $D = 1 - 8 = - 7 < 0$ , no real roots.

Q121 markMCQQuadratic Equations CBSE 2025 · Standard 30/6/3

Which of the following equations does not have a real root ?

(A) $x^{2} = 0$
(B) $2 x - 1 = 3$
(C) $x^{2} + 1 = 0$
(D) $x^{3} + x^{2} = 0$

Show answer & solution

Answer: (C)

x^{2} + 1 = 0

(A) $x^{2} = 0 \Rightarrow x = 0$ , real.
(B) $2 x - 1 = 3 \Rightarrow x = 2$ , real.
(C) $x^{2} + 1 = 0 \Rightarrow x^{2} = - 1$ ; $D = 0 - 4 = - 4 < 0$ , no real root.
(D) $x^{3} + x^{2} = x^{2} (x + 1) = 0 \Rightarrow x = 0, - 1$ , real.

Q21 markMCQQuadratic Equations CBSE 2024 · Basic 430/1/2

If the roots of quadratic equation $4 x^{2} - 5 x + k = 0$ are real and equal, then value of $k$ is :

(A) $\frac{5}{4}$
(B) $\frac{25}{16}$
(C) $- \frac{5}{4}$
(D) $- \frac{25}{16}$

Show answer & solution

Answer: (B)

\frac{25}{16}

For equal roots, $b^{2} - 4 a c = 0$ .
$25 - 16 k = 0$ , so $k = \frac{25}{16}$ .

Also asked in: 2024 Basic 430/1/3

Q31 markMCQQuadratic Equations CBSE 2024 · Basic 430/2/1

Which of the following equations has 2 as a root ?

(A) $x^{2} - 4 x + 5 = 0$
(B) $x^{2} + 3 x - 12 = 0$
(C) $2 x^{2} - 7 x + 6 = 0$
(D) $3 x^{2} - 6 x - 2 = 0$

Show answer & solution

Answer: (C)

2 x^{2} - 7 x + 6 = 0

Put $x = 2$ in each equation.
(A) $4 - 8 + 5 = 1 \neq = 0$ ; (B) $4 + 6 - 12 = - 2 \neq = 0$
(C) $8 - 14 + 6 = 0$ ; (D) $12 - 12 - 2 = - 2 \neq = 0$
So 2 is a root of $2 x^{2} - 7 x + 6 = 0$ .

Also asked in: 2024 Basic 430/2/3

Q161 markMCQQuadratic Equations CBSE 2024 · Basic 430/2/2

Which of the following quadratic equations has $- 1$ as a root ?

(A) $x^{2} - 4 x - 5 = 0$
(B) $- x^{2} - 4 x + 5 = 0$
(C) $x^{2} + 3 x + 4 = 0$
(D) $x^{2} - 5 x + 6 = 0$

Show answer & solution

Answer: (A)

x^{2} - 4 x - 5 = 0

Put $x = - 1$ :
(A) $1 + 4 - 5 = 0$ ; (B) $- 1 + 4 + 5 = 8$ ; (C) $1 - 3 + 4 = 2$ ; (D) $1 + 5 + 6 = 12$
Only (A) gives 0.

Q21 markMCQQuadratic Equations CBSE 2024 · Basic 430/3/1

The value(s) of k for which the quadratic equation $5 x^{2} - 9 k x + 5 = 0$ has real and equal roots, is/are :

(A) $\frac{- 10}{9}$
(B) $\pm \frac{9}{10}$
(C) $\frac{10}{9}$
(D) $\pm \frac{10}{9}$

Show answer & solution

Answer: (D)

\pm \frac{10}{9}

For real and equal roots, $b^{2} - 4 a c = 0$ .
$81 k^{2} - 100 = 0$ , so $k^{2} = \frac{100}{81}$ .
$k = \pm \frac{10}{9}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q81 markMCQQuadratic Equations CBSE 2024 · Basic 430/4/1

The roots of the quadratic equation $x^{2} - 4 = 0$ is/are :

(A)2 only
(B) $- 2, 2$
(C)4 only
(D) $- 4, 4$

Show answer & solution

Answer: (B)

- 2, 2

$x^{2} - 4 = 0 \Rightarrow (x - 2) (x + 2) = 0$ .
So $x = 2$ or $x = - 2$ .

Also asked in: 2024 Basic 430/4/3

Q91 markMCQQuadratic Equations CBSE 2024 · Basic 430/4/1

Which of the following is not a quadratic equation ?

(A) $(x - 2)^{2} + 1 = 2 x - 3$
(B) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(C) $x (x + 1) + 8 = (x + 2) (x - 2)$
(D) $2 x + \frac{3}{x} = 5$

Show answer & solution

Answer: (C)

x (x + 1) + 8 = (x + 2) (x - 2)

(A) $x^{2} - 4 x + 5 = 2 x - 3 \Rightarrow x^{2} - 6 x + 8 = 0$ , quadratic.
(B) $2 x^{2} - 7 x + 3 = x^{2} + 4 x - 5 \Rightarrow x^{2} - 11 x + 8 = 0$ , quadratic.
(C) $x^{2} + x + 8 = x^{2} - 4 \Rightarrow x + 12 = 0$ , which is linear, not quadratic.
(D) Multiplying by $x$ gives $2 x^{2} - 5 x + 3 = 0$ , quadratic.
So the answer is (C).

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q11 markMCQQuadratic Equations CBSE 2024 · Basic 430/4/2

The root(s) of the quadratic equation $x^{2} - 25 = 0$ is/are :

(A)5
(B) $- 5, 5$
(C)25
(D) $- 25, 25$

Show answer & solution

Answer: (B)

- 5, 5

$x^{2} - 25 = 0 \Rightarrow (x - 5) (x + 5) = 0$ .
So $x = 5$ or $x = - 5$ .

Q21 markMCQQuadratic Equations CBSE 2024 · Basic 430/5/1

The roots of the quadratic equation $x^{2} + 3 x - 10 = 0$ are :

(A)5, 2
(B) $- 5, 2$
(C) $5, - 2$
(D) $- 5, - 2$

Show answer & solution

Answer: (B)

- 5, 2

$x^{2} + 3 x - 10 = (x + 5) (x - 2)$ .
So $x = - 5$ or $x = 2$ .

Also asked in: 2024 Basic 430/5/3

Q141 markMCQQuadratic Equations CBSE 2024 · Basic 430/5/1

The discriminant of the quadratic equation $x^{2} - 4 x + 3 = 0$ is :

(A)28
(B) $- 8$
(C)4
(D)2

Show answer & solution

Answer: (C) 4

$D = b^{2} - 4 a c = (- 4)^{2} - 4 (1) (3)$ .
$D = 16 - 12 = 4$ .

Also asked in: 2024 Basic 430/5/2

Q171 markMCQQuadratic Equations CBSE 2024 · Basic 430/5/2

The roots of the quadratic equation $x^{2} + 3 x + 2 = 0$ , are :

(A) $- 1, - 2$
(B)1, 2
(C) $1, - 2$
(D) $- 1, 2$

Show answer & solution

Answer: (A)

- 1, - 2

$x^{2} + 3 x + 2 = (x + 1) (x + 2)$ .
So $x = - 1$ or $x = - 2$ .

Q141 markMCQQuadratic Equations CBSE 2024 · Basic 430/5/3

The roots of the quadratic equation $a x^{2} + b x + c = 0$ are real and distinct, if :

(A) $b^{2} - 4 a c > 0$
(B) $b^{2} - 4 a c = 0$
(C) $b^{2} - 4 a c < 0$
(D) $b^{2} - 4 a c \geq 0$

Show answer & solution

Answer: (A)

b^{2} - 4 a c > 0

The roots are real and distinct exactly when the discriminant is positive.
So $b^{2} - 4 a c > 0$ .

Q31 markMCQQuadratic Equations CBSE 2024 · Standard 30/1/1

If the roots of equation $a x^{2} + b x + c = 0, a \neq = 0$ are real and equal, then which of the following relation is true ?

(A) $a = \frac{b ^{2}}{c}$
(B) $b^{2} = a c$
(C) $a c = \frac{b ^{2}}{4}$
(D) $c = \frac{b ^{2}}{a}$

Show answer & solution

Answer: (C)

a c = \frac{b ^{2}}{4}

Real and equal roots means discriminant $b^{2} - 4 a c = 0$ .
So $b^{2} = 4 a c$ , i.e. $a c = \frac{b ^{2}}{4}$ .

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q41 markMCQQuadratic Equations CBSE 2024 · Standard 30/2/1

The quadratic equation $x^{2} + x + 1 = 0$ has ______ roots.

(A)real and equal
(B)irrational
(C)real and distinct
(D)not-real

Show answer & solution

Answer: (D) not-real

$D = b^{2} - 4 a c = 1 - 4 = - 3$
$D < 0$ , so the roots are not real.

Q131 markMCQQuadratic Equations CBSE 2024 · Standard 30/2/2

Value of k for which $x = 2$ is a solution of the equation $5 x^{2} - 4 x + (2 + k) = 0$ , is

(A)10
(B) $- 10$
(C)14
(D) $- 14$

Show answer & solution

Answer: (D)

- 14

Put $x = 2$ : $5 (4) - 4 (2) + 2 + k = 0$
$20 - 8 + 2 + k = 0$ , so $k = - 14$ .

Q81 markMCQQuadratic Equations CBSE 2024 · Standard 30/2/3

The roots of the quadratic equation $4 x^{2} - 5 x + 4 = 0$ are

(A)irrational
(B)rational and distinct
(C)not real
(D)rational and equal

Show answer & solution

Answer: (C) not real

$D = b^{2} - 4 a c = 25 - 64 = - 39$
$D < 0$ , so the roots are not real.

Q41 markMCQQuadratic Equations CBSE 2024 · Standard 30/4/1

If the discriminant of the quadratic equation $3 x^{2} - 2 x + c = 0$ is 16, then the value of $c$ is :

(A)1
(B)0
(C) $- 1$
(D) $2$

Show answer & solution

Answer: (C)

- 1

Discriminant $= b^{2} - 4 a c = (- 2)^{2} - 4 (3) (c) = 4 - 12 c$ .
$4 - 12 c = 16$ gives $- 12 c = 12$ .
$c = - 1$ .

Also asked in: 2024 Standard 30/4/2

Q171 markMCQQuadratic Equations CBSE 2024 · Standard 30/4/3

If y = 1 is one of the solutions of the quadratic equation $p y^{2} + p y + 3 = 0$ , then the value of p is :

(A) $- 3$
(B)2
(C) $- \frac{3}{2}$
(D) $- 2$

Show answer & solution

Answer: (C)

- \frac{3}{2}

Put $y = 1$ : $p + p + 3 = 0$ .
$2 p = - 3$ , so $p = - \frac{3}{2}$ .

Q61 markMCQQuadratic Equations CBSE 2024 · Standard 30/5/2

If the quadratic equation $a x^{2} + b x + c = 0$ has real and equal roots, then the value of $c$ is :

(A) $\frac{b}{2 a}$
(B) $- \frac{b}{2 a}$
(C) $\frac{b ^{2}}{4 a}$
(D) $- \frac{b ^{2}}{4 a}$

Show answer & solution

Answer: (C)

\frac{b ^{2}}{4 a}

Equal roots means $b^{2} - 4 a c = 0$ .
So $c = \frac{b ^{2}}{4 a}$ .

Q41 markMCQQuadratic Equations CBSE 2023 · Basic 430/1/1

The discriminant of the quadratic equation $2 x^{2} - 5 x - 3 = 0$ is

(A)1
(B)49
(C)7
(D)19

Show answer & solution

Answer: (B) 49

Here $a = 2$ , $b = - 5$ , $c = - 3$ .
$D = b^{2} - 4 a c = 25 - 4 (2) (- 3) = 25 + 24 = 49$

Also asked in: 2023 Basic 430/1/3

Q61 markMCQQuadratic Equations CBSE 2023 · Basic 430/1/2

The discriminant of the quadratic equation $2 x^{2} + x - 1 = 0$ is :

(A) $- 9$
(B) $- 7$
(C)9
(D)7

Show answer & solution

Answer: (C) 9

Here $a = 2$ , $b = 1$ , $c = - 1$ .
$D = b^{2} - 4 a c = 1 - 4 (2) (- 1) = 1 + 8 = 9$

Q101 markMCQQuadratic Equations CBSE 2023 · Basic 430/2/1

If the quadratic equation $9 x^{2} + b x + \frac{1}{4} = 0$ has equal roots, then the value of b is :

(A)0
(B) $- 3$ only
(C)3 only
(D) $\pm 3$

Show answer & solution

Answer: (d)

\pm 3

For equal roots, $b^{2} - 4 a c = 0$ .
$b^{2} - 4 \times 9 \times \frac{1}{4} = 0$
$b^{2} = 9$ , so $b = \pm 3$

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q151 markMCQQuadratic Equations CBSE 2023 · Basic 430/4/1

A quadratic equation whose one root is 2 and the sum of whose roots is zero, is :

(A) $x^{2} + 4 = 0$
(B) $x^{2} - 2 = 0$
(C) $4 x^{2} - 1 = 0$
(D) $x^{2} - 4 = 0$

Show answer & solution

Answer: (D)

x^{2} - 4 = 0

Sum of roots is 0 and one root is 2, so the other root is $- 2$ .
Equation: $(x - 2) (x + 2) = 0$ , i.e. $x^{2} - 4 = 0$ .

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q161 markMCQQuadratic Equations CBSE 2023 · Basic 430/4/1

Which of the following is not a quadratic equation ?

(A) $2 (x - 1)^{2} = 4 x^{2} - 2 x + 1$
(B) $2 x - x^{2} = x^{2} + 5$
(C) $(2 x + 3)^{2} + x^{2} = 3 x^{2} - 5 x$
(D) $(x^{2} + 2 x)^{2} = x^{4} + 3 + 4 x^{3}$

Show answer & solution

Answer: (C)

(2 x + 3)^{2} + x^{2} = 3 x^{2} - 5 x

(A): $2 x^{2} - 4 x + 2 = 4 x^{2} - 2 x + 1 \Rightarrow 2 x^{2} + 2 x - 1 = 0$ , quadratic.
(B): $2 x^{2} - 2 x + 5 = 0$ , quadratic.
(C): $2 x^{2} + 26 x + 3 + x^{2} = 3 x^{2} - 5 x \Rightarrow (26 + 5) x + 3 = 0$ , which is linear.
(D): $x^{4} + 4 x^{3} + 4 x^{2} = x^{4} + 4 x^{3} + 3 \Rightarrow 4 x^{2} - 3 = 0$ , quadratic.
So (C) is not quadratic.

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q201 markAssertion–ReasonQuadratic Equations CBSE 2023 · Basic 430/4/1

Assertion (A) : If one root of the quadratic equation $4 x^{2} - 10 x + (k - 4) = 0$ is reciprocal of the other, then value of k is 8.
Reason (R) : Roots of the quadratic equation $x^{2} - x + 1 = 0$ are real.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) gives the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true but Reason (R) does not give the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (C) A is true but R is false.

If the roots are $α$ and $\frac{1}{α}$ , their product is 1, so $\frac{k - 4}{4} = 1$ , giving $k = 8$ . A is true.
For $x^{2} - x + 1 = 0$ , $D = 1 - 4 = - 3 < 0$ , so the roots are not real. R is false.

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q51 markMCQQuadratic Equations CBSE 2023 · Basic 430/5/1

The value(s) of k for which the roots of quadratic equation $x^{2} + 4 x + k = 0$ are real, is :

(A) $k \geq 4$
(B) $k \leq 4$
(C) $k \geq - 4$
(D) $k \leq - 4$

Show answer & solution

Answer: (B)

k \leq 4

For real roots, $b^{2} - 4 a c \geq 0$ .
$16 - 4 k \geq 0$ , so $k \leq 4$ .

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q101 markMCQQuadratic Equations CBSE 2023 · Basic 430/5/1

The two roots of the equation $3 x^{2} - 26 x + 2 = 0$ are :

(A)real and distinct
(B)not real
(C)real and equal
(D)rational

Show answer & solution

Answer: (C) real and equal

$D = b^{2} - 4 a c = (26)^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$ .
So the roots are real and equal.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q81 markMCQQuadratic Equations CBSE 2023 · Standard 30/1/1

The least positive value of k, for which the quadratic equation $2 x^{2} + k x - 4 = 0$ has rational roots, is

(A) $\pm 22$
(B)2
(C) $\pm 2$
(D) $2$

Show answer & solution

Answer: (B) 2

Roots are rational when the discriminant is a perfect square.
$D = k^{2} - 4 (2) (- 4) = k^{2} + 32$ .
k = 1 gives D = 33, not a perfect square.
k = 2 gives $D = 36 = 6^{2}$ , a perfect square (roots 1 and –2).
Least positive value of k is 2.

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q11 markMCQQuadratic Equations CBSE 2023 · Standard 30/2/1

Which of the following quadratic equations has sum of its roots as 4 ?

(A) $2 x^{2} - 4 x + 8 = 0$
(B) $- x^{2} + 4 x + 4 = 0$
(C) $2 x^{2} - \frac{4}{2} x + 1 = 0$
(D) $4 x^{2} - 4 x + 4 = 0$

Show answer & solution

Answer: (B)

- x^{2} + 4 x + 4 = 0

Sum of roots of $a x^{2} + b x + c = 0$ is $- \frac{b}{a}$ .
(A) $- \frac{- 4}{2} = 2$ ; (B) $- \frac{4}{- 1} = 4$ ; (C) $\frac{4/ 2}{2} = 2$ ; (D) $\frac{4}{4} = 1$ .
Only (B) gives sum 4.

Also asked in: 2023 Standard 30/2/3

Q51 markMCQQuadratic Equations CBSE 2023 · Standard 30/2/2

If $x = 0.3$ , is a root of the equation $x^{2} - 0.9 k = 0$ , then k is equal to :

(A)1
(B)10
(C)0.1
(D)100

Show answer & solution

Answer: (C) 0.1

Substitute $x = 0.3$ : $0.09 - 0.9 k = 0$
$k = \frac{0.09}{0.9} = 0.1$

Q21 markMCQQuadratic Equations CBSE 2023 · Standard 30/4/1

The roots of the equation $x^{2} + 3 x - 10 = 0$ are :

(A)2, $- 5$
(B) $- 2$ , 5
(C)2, 5
(D) $- 2$ , $- 5$

Show answer & solution

Answer: (A) 2,

- 5

$x^{2} + 3 x - 10 = x^{2} + 5 x - 2 x - 10 = (x + 5) (x - 2)$
$(x + 5) (x - 2) = 0 \Rightarrow x = 2$ or $x = - 5$

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q51 markMCQQuadratic Equations CBSE 2023 · Standard 30/5/1

A quadratic equation whose roots are $(2 + 3)$ and $(2 - 3)$ is :

(A) $x^{2} - 4 x + 1 = 0$
(B) $x^{2} + 4 x + 1 = 0$
(C) $4 x^{2} - 3 = 0$
(D) $x^{2} - 1 = 0$

Show answer & solution

Answer: (A)

x^{2} - 4 x + 1 = 0

Sum of roots $= (2 + 3) + (2 - 3) = 4$ .
Product of roots $= (2 + 3) (2 - 3) = 4 - 3 = 1$ .
Equation: $x^{2} - (sum) x + product = 0$ , i.e. $x^{2} - 4 x + 1 = 0$ .

Also asked in: 2023 Standard 30/5/2

Q101 markMCQQuadratic Equations CBSE 2023 · Standard 30/5/3

A quadratic equation whose roots are $(3 - 2)$ and $(3 + 2)$ is :

(A) $x^{2} - 6 x + 7 = 0$
(B) $x^{2} + 6 x + 7 = 0$
(C) $9 x^{2} - 2 = 0$
(D) $x^{2} - 7 = 0$

Show answer & solution

Answer: (A)

x^{2} - 6 x + 7 = 0

Sum of roots $= (3 - 2) + (3 + 2) = 6$ .
Product of roots $= 9 - 2 = 7$ .
Equation: $x^{2} - 6 x + 7 = 0$ .

Q91 markMCQQuadratic Equations CBSE 2023 · Standard 30/6/1

If the quadratic equation $a x^{2} + b x + c = 0$ has two real and equal roots, then 'c' is equal to

(A) $\frac{- b}{2 a}$
(B) $\frac{b}{2 a}$
(C) $\frac{- b ^{2}}{4 a}$
(D) $\frac{b ^{2}}{4 a}$

Show answer & solution

Answer: (D)

\frac{b ^{2}}{4 a}

Equal roots means $b^{2} - 4 a c = 0$ .
So $c = \frac{b ^{2}}{4 a}$ .

Also asked in: 2023 Standard 30/6/2

Q191 markAssertion–ReasonQuadratic Equations CBSE 2023 · Standard 30/6/1

Statement A (Assertion) : If $5 + 7$ is a root of a quadratic equation with rational co-efficients, then its other root is $5 - 7$ .
Statement R (Reason) : Surd roots of a quadratic equation with rational co-efficients occur in conjugate pairs.

(A)Both Assertion (A) and Reason (R) are true; and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true; but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true but Reason (R) is false.
(D)Assertion (A) is false but Reason (R) is true.

Show answer & solution

Answer: (A) Both A and R are true and R is the correct explanation of A.

By the quadratic formula, roots with rational coefficients are $\frac{- b \pm D}{2 a}$ , so irrational (surd) roots occur in conjugate pairs. R is true.
Hence if $5 + 7$ is a root, $5 - 7$ is the other root. A is true and follows from R.

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q161 markMCQQuadratic Equations CBSE 2023 · Standard 30/6/3

If 'p' is a root of the quadratic equation $x^{2} - (p + q) x + k = 0$ , then the value of 'k' is

(A)p
(B)q
(C)p + q
(D)pq

Show answer & solution

Answer: (D) pq

Put x = p: $p^{2} - (p + q) p + k = 0$ .
$p^{2} - p^{2} - pq + k = 0$ , so k = pq.

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