CBSE Maths › Class 10 PYQs › Quadratic Equations

Quadratic Equations: 2 marks Questions (CBSE Class 10)

38 different 2 marks 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)

Q212 marksVery Short AnswerQuadratic Equations CBSE 2026 · Standard 30/4/1

Verify that roots of the quadratic equation $(p - q) x^{2} + (q - r) x + (r - p) = 0$ are equal when $q + r = 2 p$ .

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Answer: Proved.

$D = (q - r)^{2} - 4 (p - q) (r - p)$
Given $q + r = 2 p$ , so $p = \frac{q + r}{2}$ .
Then $p - q = \frac{r - q}{2}$ and $r - p = \frac{r - q}{2}$ .
$D = (q - r)^{2} - 4 \times \frac{( r - q ) ^{2}}{4} = (q - r)^{2} - (r - q)^{2} = 0$
Since $D = 0$ , the roots are equal.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q212 marksVery Short AnswerQuadratic Equations CBSE 2025 · Basic 430/6/1

Solve the equation $4 x^{2} - 9 x + 3 = 0$ , using quadratic formula.

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Answer:

x = \frac{9 + 33}{8}

\frac{9 - 33}{8}

$a = 4$ , $b = - 9$ , $c = 3$ .
$D = b^{2} - 4 a c = 81 - 48 = 33$ .
$x = \frac{- b \pm D}{2 a} = \frac{9 \pm 33}{8}$ .

Also asked in: 2025 Basic 430/6/2

Q21 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2025 · Basic 430/6/1

Find the nature of roots of the equation $3 x^{2} - 43 x + 4 = 0$ .

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Answer: Real and equal roots

$a = 3$ , $b = - 43$ , $c = 4$ .
$D = b^{2} - 4 a c = 48 - 48 = 0$ .
Since D = 0, the roots are real and equal.

Also asked in: 2025 Basic 430/6/2

Q232 marksVery Short AnswerQuadratic Equations CBSE 2025 · Basic 430/6/3

Solve the quadratic equation $3 x^{2} + 10 x + 73 = 0$ using quadratic formula.

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Answer:

x = - 3

- \frac{7}{3}

(i.e.

- \frac{7 3}{3}

)

$a = 3$ , $b = 10$ , $c = 73$ .
$D = 100 - 4 \times 3 \times 73 = 100 - 84 = 16$ .
$x = \frac{- 10 \pm 4}{2 3}$ .
$x = \frac{- 6}{2 3} = - 3$ or $x = \frac{- 14}{2 3} = - \frac{7}{3}$ .

Q23 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2025 · Basic 430/6/3

Find the nature of roots of the equation $4 x^{2} - 4 a^{2} x + a^{4} - b^{4} = 0$ , $b \neq = 0$

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Answer: Real and distinct roots

$D = (- 4 a^{2})^{2} - 4 \times 4 \times (a^{4} - b^{4}) = 16 a^{4} - 16 a^{4} + 16 b^{4} = 16 b^{4}$ .
Since $b \neq = 0$ , $D = 16 b^{4} > 0$ .
So the roots are real and distinct.

Q242 marksVery Short AnswerQuadratic Equations CBSE 2025 · Standard 30/3/1

Find the value(s) of 'k' so that the quadratic equation $4 x^{2} + k x + 1 = 0$ has real and equal roots.

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Answer:

k = \pm 4

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$k^{2} - 4 (4) (1) = 0$ , so $k^{2} = 16$ .
$k = 4$ or $k = - 4$ .

Also asked in: 2025 Standard 30/3/3

Q232 marksVery Short AnswerQuadratic Equations CBSE 2024 · Basic 430/5/1

If $\frac{2}{3}$ is a root of the quadratic equation $k x^{2} - x - 2 = 0$ , then find the value of k.

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Answer:

k = 6

Put $x = \frac{2}{3}$ : $k \times \frac{4}{9} - \frac{2}{3} - 2 = 0$ .
$\frac{4 k}{9} = \frac{8}{3}$ .
$k = \frac{8}{3} \times \frac{9}{4} = 6$ .

Also asked in: 2024 Basic 430/5/3

Q212 marksVery Short AnswerQuadratic Equations CBSE 2024 · Basic 430/5/2

If $\frac{1}{2}$ is a root of the quadratic equation $x^{2} + k x - \frac{5}{4} = 0$ , then find the value of k.

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Answer:

k = 2

Put $x = \frac{1}{2}$ : $\frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0$ .
$\frac{k}{2} = 1$ , so $k = 2$ .

Q232 marksVery Short AnswerQuadratic Equations CBSE 2023 · Basic 430/4/1

Find the discriminant of the quadratic equation $3 x^{2} - 2 x + \frac{1}{3} = 0$ and hence find the nature of its roots.

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Answer:

D = 0

; two equal real roots.

$a = 3$ , $b = - 2$ , $c = \frac{1}{3}$ .
$D = b^{2} - 4 a c = 4 - 4 \times 3 \times \frac{1}{3} = 4 - 4 = 0$ .
Since $D = 0$ , the equation has two equal real roots.

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

Q23 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2023 · Basic 430/4/1

Find the roots of the quadratic equation $x^{2} - x - 2 = 0$ .

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Answer:

x = 2

x = - 1

$x^{2} - x - 2 = x^{2} - 2 x + x - 2 = (x - 2) (x + 1)$ .
$(x - 2) (x + 1) = 0 \Rightarrow x = 2$ or $x = - 1$ .

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

Q232 marksVery Short AnswerQuadratic Equations CBSE 2023 · Basic 430/6/1

Find the value of k for which the roots of the quadratic equation $5 x^{2} - 10 x + k = 0$ are real and equal.

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Answer:

k = 5

For real and equal roots, $b^{2} - 4 a c = 0$ .
$(- 10)^{2} - 4 (5) (k) = 0$ , so $100 - 20 k = 0$ .
$k = 5$ .

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q23 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2023 · Basic 430/6/1

If one root of the quadratic equation $3 x^{2} - 8 x - (2 k + 1) = 0$ is seven times the other, then find the value of k.

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Answer:

k = - \frac{5}{3}

Let the roots be $α$ and $7 α$ .
Sum: $8 α = \frac{8}{3}$ , so $α = \frac{1}{3}$ .
Product: $7 α^{2} = \frac{- ( 2 k + 1 )}{3}$ , so $\frac{7}{9} = \frac{- ( 2 k + 1 )}{3}$ .
$2 k + 1 = - \frac{7}{3}$ , so $2 k = - \frac{10}{3}$ and $k = - \frac{5}{3}$ .

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q232 marksVery Short AnswerQuadratic Equations CBSE 2023 · Standard 30/4/1

Find the sum and product of the roots of the quadratic equation $2 x^{2} - 9 x + 4 = 0$ .

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Answer: Sum

= \frac{9}{2}

, product = 2

Here $a = 2$ , $b = - 9$ , $c = 4$ .
Sum of roots $= - \frac{b}{a} = \frac{9}{2}$
Product of roots $= \frac{c}{a} = \frac{4}{2} = 2$

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

Q23 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2023 · Standard 30/4/1

Find the discriminant of the quadratic equation $4 x^{2} - 5 = 0$ and hence comment on the nature of roots of the equation.

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Answer: Discriminant = 80; the roots are real and distinct.

Here $a = 4$ , $b = 0$ , $c = - 5$ .
$D = b^{2} - 4 a c = 0 - 4 (4) (- 5) = 80$
Since $D > 0$ , the equation has two distinct real roots.

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

Q32 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/1/1

Find the nature of the roots of the quadratic equation $x^{2} - 5 x + 9 = 0$ .

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Answer: No real roots (discriminant

= - 11 < 0

)

Here $a = 1$ , $b = - 5$ , $c = 9$
$D = b^{2} - 4 a c = 25 - 36 = - 11$
Since $D < 0$ , the equation has no real roots.

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q3 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/1/1

Write a quadratic equation with roots $- 3$ and 5.

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Answer:

x^{2} - 2 x - 15 = 0

Sum of roots $= - 3 + 5 = 2$
Product of roots $= (- 3) (5) = - 15$
Required equation: $x^{2} - (sum) x + product = 0$
$x^{2} - 2 x - 15 = 0$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q52 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/1/1

Solve the quadratic equation $2 x^{2} - 5 x - 1 = 0$ for x.

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Answer:

x = \frac{5 + 33}{4}

x = \frac{5 - 33}{4}

$a = 2$ , $b = - 5$ , $c = - 1$
$D = b^{2} - 4 a c = 25 + 8 = 33$
$x = \frac{- b \pm D}{2 a} = \frac{5 \pm 33}{4}$

Also asked in: 2022 Basic 430/1/3

Q42 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/1/2

Solve the quadratic equation $x^{2} + 3 x - 9 = 0$ for x.

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Answer:

x = \frac{- 3 + 3 5}{2}

x = \frac{- 3 - 3 5}{2}

$a = 1$ , $b = 3$ , $c = - 9$
$D = 9 + 36 = 45$ , $D = 35$
$x = \frac{- 3 \pm 3 5}{2}$

Q12 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/2/1

Find the nature of the roots of the quadratic equation :
$4 x^{2} - 5 x - 1 = 0$

Show answer & solution

Answer: The roots are real and distinct (unequal), since

D = 41 > 0

Here $a = 4$ , $b = - 5$ , $c = - 1$ .
$D = b^{2} - 4 a c = 25 - 4 (4) (- 1) = 25 + 16 = 41$ .
Since $D > 0$ , the equation has two distinct real roots.
(As 41 is not a perfect square, the roots are irrational.)

Also asked in: 2022 Basic 430/2/3

Q52 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/2/2

Solve the equation : $3 x^{2} - 8 x - 1 = 0$ for $x$ .

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Answer:

x = \frac{4 + 19}{3}

x = \frac{4 - 19}{3}

$a = 3$ , $b = - 8$ , $c = - 1$ .
$D = 64 - 4 (3) (- 1) = 64 + 12 = 76$ .
$x = \frac{8 \pm 76}{6} = \frac{8 \pm 2 19}{6} = \frac{4 \pm 19}{3}$ .

Q12 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/3/1

For what value of p, does the quadratic equation $p x^{2} + 2 x + p = 0$ have real and equal roots ?

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Answer:

p = 1

p = - 1

For real and equal roots, the discriminant must be zero.
$D = b^{2} - 4 a c = 2^{2} - 4 (p) (p) = 4 - 4 p^{2}$
$4 - 4 p^{2} = 0 \Rightarrow p^{2} = 1$
So $p = 1$ or $p = - 1$ (both non-zero, so the equation stays quadratic).

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q1 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/3/1

Solve the quadratic equation for x : $6 - x - x^{2} = 0$

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Answer:

x = 2

x = - 3

Multiply by $- 1$ : $x^{2} + x - 6 = 0$
Split the middle term: $x^{2} + 3 x - 2 x - 6 = 0$
$x (x + 3) - 2 (x + 3) = 0 \Rightarrow (x + 3) (x - 2) = 0$
So $x = 2$ or $x = - 3$ .

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q12 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/4/1

Find the value of 'k' so that the quadratic equation $3 x^{2} - 5 x - 2 k = 0$ has real and equal roots.

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Answer:

k = - \frac{25}{24}

Here $a = 3$ , $b = - 5$ , $c = - 2 k$ .
For real and equal roots, $b^{2} - 4 a c = 0$ .
$25 - 4 (3) (- 2 k) = 0 \Rightarrow 25 + 24 k = 0$ .
So $k = - \frac{25}{24}$ .

Q62 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/4/2

Find the value(s) of 'a' for which the quadratic equation $x^{2} - a x + 1 = 0$ has real and equal roots.

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Answer:

a = 2

a = - 2

For real and equal roots, discriminant $= 0$ .
$(- a)^{2} - 4 (1) (1) = 0 \Rightarrow a^{2} = 4$ .
$a = \pm 2$ .

Q52 marksVery Short AnswerQuadratic Equations CBSE 2022 · Basic 430/4/3

Solve for $x$ : $2 x^{2} + \frac{7}{2} x + \frac{3}{4} = 0$ .

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Answer:

x = - \frac{1}{4}

x = - \frac{3}{2}

Multiply by 4: $8 x^{2} + 14 x + 3 = 0$ .
$8 x^{2} + 12 x + 2 x + 3 = 0 \Rightarrow 4 x (2 x + 3) + 1 (2 x + 3) = 0$ .
$(4 x + 1) (2 x + 3) = 0$ .
$x = - \frac{1}{4}$ or $x = - \frac{3}{2}$ .

Q32 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/1/1

Find the value of $m$ for which the quadratic equation
$(m - 1) x^{2} + 2 (m - 1) x + 1 = 0$
has two real and equal roots.

Show answer & solution

Answer:

m = 2

For equal roots, $D = b^{2} - 4 a c = 0$ (and $m - 1 \neq = 0$ ).
$[2 (m - 1)]^{2} - 4 (m - 1) (1) = 0$
$4 (m - 1) (m - 1 - 1) = 0 \Rightarrow 4 (m - 1) (m - 2) = 0$
$m = 1$ is rejected since then the equation is not quadratic.
So $m = 2$ .

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q3 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/1/1

Solve the following quadratic equation for x :
$3 x^{2} + 10 x + 73 = 0$

Show answer & solution

Answer:

x = - 3

x = - \frac{7}{3} = - \frac{7 3}{3}

Split the middle term: $3 \times 73 = 21 = 3 \times 7$ and $3 + 7 = 10$ .
$3 x^{2} + 3 x + 7 x + 73 = 0$
$3 x (x + 3) + 7 (x + 3) = 0$
$(x + 3) (3 x + 7) = 0$
$x = - 3$ or $x = - \frac{7}{3} = - \frac{7 3}{3}$

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q52 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/1/1

The product of Rehan’s age (in years) 5 years ago and his age 7 years from now, is one more than twice his present age. Find his present age.

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Answer: 6 years

Let the present age be $x$ years.
$(x - 5) (x + 7) = 2 x + 1$
$x^{2} + 2 x - 35 = 2 x + 1 \Rightarrow x^{2} = 36$
$x = 6$ (age cannot be negative).
Rehan's present age is 6 years.

Also asked in: 2022 Standard 30/1/2, 2022 Standard 30/1/3

Q12 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/2/1

Solve the quadratic equation : $x^{2} + 22 x - 6 = 0$ for $x$ .

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Answer:

x = 2

x = - 32

Here $a = 1$ , $b = 22$ , $c = - 6$ .
$D = b^{2} - 4 a c = 8 + 24 = 32$ , so $D = 42$ .
$x = \frac{- 2 2 \pm 4 2}{2}$ .
$x = \frac{2 2}{2} = 2$ or $x = \frac{- 6 2}{2} = - 32$ .
(Check by factorising: $(x - 2) (x + 32) = x^{2} + 22 x - 6$ .)

Also asked in: 2022 Standard 30/2/3

Q52 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/2/2

Solve the quadratic equation : $x^{2} - 2 a x + (a^{2} - b^{2}) = 0$ for $x$ .

Show answer & solution

Answer:

x = a + b

x = a - b

$x^{2} - 2 a x + a^{2} - b^{2} = 0$
$(x - a)^{2} - b^{2} = 0$
$(x - a - b) (x - a + b) = 0$
$x = a + b$ or $x = a - b$ .

Q12 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/3/1

Solve the quadratic equation for x :
$x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$

Show answer & solution

Answer:

x = a + 2 b

x = a - 2 b

Here $A = 1$ , $B = - 2 a$ , $C = - (4 b^{2} - a^{2})$ .
$D = B^{2} - 4 A C = 4 a^{2} + 4 (4 b^{2} - a^{2}) = 16 b^{2}$ .
$x = \frac{2 a \pm 16 b ^{2}}{2} = \frac{2 a \pm 4 b}{2} = a \pm 2 b$ .
So $x = a + 2 b$ or $x = a - 2 b$ .

Also asked in: 2022 Standard 30/3/3

Q1 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/3/1

If the quadratic equation
$(1 + a^{2}) x^{2} + 2 ab x + (b^{2} - c^{2}) = 0$
has equal and real roots, then prove that :
$b^{2} = c^{2} (1 + a^{2})$

Show answer & solution

Answer: Proved.

For equal roots, $D = 0$ : $(2 ab)^{2} - 4 (1 + a^{2}) (b^{2} - c^{2}) = 0$ .
$\Rightarrow a^{2} b^{2} - (b^{2} - c^{2} + a^{2} b^{2} - a^{2} c^{2}) = 0$
$\Rightarrow - b^{2} + c^{2} + a^{2} c^{2} = 0$
$\Rightarrow b^{2} = c^{2} (1 + a^{2})$ . Hence proved.

Also asked in: 2022 Standard 30/3/3

Q42 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/3/2

For what value of m, the quadratic equation
$m x^{2} - 2 (m - 1) x + (m + 2) = 0$
has two real and equal roots ?

Show answer & solution

Answer:

m = \frac{1}{4}

For real and equal roots, $D = 0$ : $[- 2 (m - 1)]^{2} - 4 m (m + 2) = 0$ .
$4 (m^{2} - 2 m + 1) - 4 m^{2} - 8 m = 0 \Rightarrow - 16 m + 4 = 0$ .
$m = \frac{1}{4}$ (and $m \neq = 0$ , so the equation is quadratic).

Q4 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/3/2

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

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Answer: Shorter side 90 m, longer side 120 m

Let the shorter side be x m. Then longer side $= x + 30$ and diagonal $= x + 60$ .
$x^{2} + (x + 30)^{2} = (x + 60)^{2} \Rightarrow 2 x^{2} + 60 x + 900 = x^{2} + 120 x + 3600$ .
$x^{2} - 60 x - 2700 = 0 \Rightarrow (x - 90) (x + 30) = 0 \Rightarrow x = 90$ (x cannot be negative).
Sides are 90 m and 120 m.

Q32 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/4/1

If the sum of the roots of the quadratic equation $k y^{2} - 11 y + (k - 23) = 0$ is $\frac{13}{21}$ more than the product of the roots, then find the value of $k$ .

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Answer:

k = 21

Sum of roots $= \frac{11}{k}$ , product of roots $= \frac{k - 23}{k}$ .
Given: $\frac{11}{k} - \frac{k - 23}{k} = \frac{13}{21}$
$\frac{34 - k}{k} = \frac{13}{21}$
$21 (34 - k) = 13 k \Rightarrow 714 = 34 k$
$k = 21$

Also asked in: 2022 Standard 30/4/2

Q3 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/4/1

If $x = - 2$ is the common solution of quadratic equations $a x^{2} + x - 3 a = 0$ and $x^{2} + b x + b = 0$ , then find the value of $a^{2} b$ .

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Answer:

a^{2} b = 16

Put $x = - 2$ in $a x^{2} + x - 3 a = 0$ : $4 a - 2 - 3 a = 0 \Rightarrow a = 2$ .
Put $x = - 2$ in $x^{2} + b x + b = 0$ : $4 - 2 b + b = 0 \Rightarrow b = 4$ .
$a^{2} b = 2^{2} \times 4 = 16$

Also asked in: 2022 Standard 30/4/2

Q12 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/4/3

Find the value of ‘k’ for which the quadratic equation $2 k x^{2} - 40 x + 25 = 0$ has real and equal roots.

Show answer & solution

Answer:

k = 8

For real and equal roots, $D = b^{2} - 4 a c = 0$ .
$(- 40)^{2} - 4 (2 k) (25) = 0$
$1600 - 200 k = 0$
$k = 8$

Q1 (OR) (OR)2 marksVery Short AnswerQuadratic Equations CBSE 2022 · Standard 30/4/3

Solve for $x$ : $\frac{5}{2} x^{2} + \frac{2}{5} = 1 - 2 x$ .

Show answer & solution

Answer:

x = \frac{- 2 + 10}{5}

x = \frac{- 2 - 10}{5}

Multiply by 10: $25 x^{2} + 4 = 10 - 20 x$
$25 x^{2} + 20 x - 6 = 0$
$D = 2 0^{2} - 4 (25) (- 6) = 400 + 600 = 1000$
$x = \frac{- 20 \pm 1000}{50} = \frac{- 20 \pm 10 10}{50}$
$x = \frac{- 2 \pm 10}{5}$

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