Class 10 Quadratic Equations — practice questions with answers

Algebra — Quadratics · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. Solve x² − 7x + 10 = 0.

A2 and 5
B1 and 10
C−2 and −5
D3 and 4

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Answer: A) 2 and 5

Product 10, sum 7 → numbers 2 and 5. (x−2)(x−5)=0.

2. Assertion (A): The equation x² + 1 = 0 has no real roots. Reason (R): Its discriminant is negative. Which is correct?

ABoth A and R are true, and R is the correct explanation of A
BBoth A and R are true, but R is not the correct explanation of A
CA is true but R is false
DA is false but R is true

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

For x² + 1 = 0, D = 0 − 4(1)(1) = −4 < 0, so there are no real roots. The negative discriminant is exactly the reason, so R correctly explains A.

3. Assertion (A): x² − 4x + 4 = 0 has two equal roots. Reason (R): A quadratic with discriminant zero has coincident roots. Which is correct?

ABoth A and R are true, and R is the correct explanation of A
BBoth A and R are true, but R is not the correct explanation of A
CA is true but R is false
DA is false but R is true

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

For x² − 4x + 4 = 0, D = 16 − 16 = 0, giving the equal roots x = 2, 2. R states the general rule and correctly explains A.

4. Arun applies the quadratic formula to 2x² − 3x + 4 = 0. What does he find?

AD = −23, so $D$ is not a real number and the equation has no real roots
BD = 23, so the roots are $\frac{3 \pm 23}{4}$
CD = 41, so the roots are $\frac{3 \pm 41}{4}$
DD = 0, so there is one repeated root $x = \frac{3}{4}$

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Answer: A) D = −23, so

D

is not a real number and the equation has no real roots

D = (−3)² − 4(2)(4) = 9 − 32 = −23. A negative discriminant means

D

has no real value, so the formula produces no real roots and Arun should stop there. Reading D as +23 flips the sign of the negative result — the single most common discriminant error. D = 41 comes from 9 + 32, i.e. adding 4ac instead of subtracting it. D = 0 would require b² = 4ac, which is false here.

5. Ananya thinks of a positive integer. Adding the number to its reciprocal gives her

\frac{50}{7}

. Which integer did she think of?

A50
B14
C $\frac{1}{7}$
D7

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Answer: D) 7

Let the number be x:

x + \frac{1}{x} = \frac{50}{7}

. Multiply by 7x: 7x² + 7 = 50x → 7x² − 50x + 7 = 0. D = 2500 − 196 = 2304 = 48², so

x = \frac{50 \pm 48}{14}

= 7 or

\frac{1}{7}

. Both are positive, but only 7 is an INTEGER, so that is Ananya's number (7 + 1/7 = 50/7).

\frac{1}{7}

is the other root, correctly rejected here by the word "integer" rather than by sign. 50 and 14 are the numerator and denominator of the working, not roots.

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