Class 10 Maths Chapter 4 Quadratic Equations Worksheet with Answers, free PDF (based on NCERT)

A 20-mark worksheet on Quadratic Equations for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

Which of the following is NOT a quadratic equation?

(a) (x + 1)³ = x³ + 2(b) (x − 2)² + 1 = 2x − 3(c) x(2x + 3) = x² + 1(d) (x + 1)³ = 2x(x² + 1)

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Answer: (d) (x + 1)³ = 2x(x² + 1)

In (d) the left side has x³ and the right side 2x³; they do not cancel, and the equation becomes −x³ + 3x² + x + 1 = 0, a cubic. The others reduce to 3x² + 3x − 1 = 0, x² − 6x + 8 = 0 and x² + 3x − 1 = 0.

Question 2

The solution of x²x − 3 = 9x − 3 is

(a) x = 3 only(b) x = −3 only(c) x = 3 and x = −3(d) no value of x

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Answer: (b) x = −3 only

Multiplying by (x − 3) gives x² = 9, so x = 3 or x = −3. But x = 3 makes the denominator zero in the given equation, so it is rejected; the solution is x = −3.

Question 3

The roots of (3x − 2)(x + 4) = 0 are

(a) 2 and −4(b) −23 and 4(c) 32 and −4(d) 23 and −4

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Answer: (d) 23 and −4

Set each factor to zero: 3x − 2 = 0 gives x = 2/3; x + 4 = 0 gives x = −4.

Question 4

When (x − 1)³ = x³ + 2 is simplified, it becomes

(a) x² + x + 1 = 0(b) x² − x + 1 = 0(c) x² − x − 1 = 0(d) x³ − 3x² + 3x − 3 = 0

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Answer: (b) x² − x + 1 = 0

x³ − 3x² + 3x − 1 = x³ + 2 gives −3x² + 3x − 3 = 0; dividing by −3, x² − x + 1 = 0. The x³ terms cancel, so it is quadratic.

Question 5

If x = 2 is a root of x² + kx − 10 = 0, then k is

(a) 3(b) −3(c) 7(d) 2

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Answer: (a) 3

Substitute x = 2: 4 + 2k − 10 = 0, so 2k = 6 and k = 3.

Question 6 · 2 marks

Check whether (2x + 1)(x − 3) = (x + 4)(x − 2) is a quadratic equation.

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LHS = 2x² − 5x − 3; RHS = x² + 2x − 8. Bringing all terms to one side: x² − 7x + 5 = 0. This is of the form ax² + bx + c = 0 with a = 1 ≠ 0, so it is a quadratic equation.

expands both sides correctly (1)
reduces to x² − 7x + 5 = 0 and concludes it is quadratic since a ≠ 0 (1)

Question 7 · 2 marks

Find the roots of 2x² − 2√2x + 1 = 0 by factorisation.

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Split −2√2x as −√2x − √2x (product 2 = ac, sum −2√2 = b). 2x² − √2x − √2x + 1 = √2x(√2x − 1) − 1(√2x − 1) = (√2x − 1)(√2x − 1) = 0. So x = 1/√2, 1/√2 (a repeated root).

splits the middle term and factorises to (√2x − 1)(√2x − 1) (1)
roots 1/√2 and 1/√2, i.e. √2/2 repeated (1)

Question 8 · 2 marks

Find the discriminant of 2x² − 4x + 3 = 0 and hence state the nature of its roots.

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a = 2, b = −4, c = 3. D = b² − 4ac = 16 − 24 = −8. Since D < 0, the equation has no real roots.

D = −8 (1)
no real roots, because D < 0 (1)

Question 9 · 3 marks

Solve 1x − 1 + 1x + 3 = 23, x ≠ 1, −3.

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Combine: (x + 3 + x − 1)/((x − 1)(x + 3)) = 2/3, so 3(2x + 2) = 2(x² + 2x − 3). 6x + 6 = 2x² + 4x − 6 gives 2x² − 2x − 12 = 0, i.e. x² − x − 6 = 0, (x − 3)(x + 2) = 0. x = 3 or x = −2; neither makes a denominator zero. Check x = 3: 1/2 + 1/6 = 2/3.

clears the denominators correctly (1)
reduces to x² − x − 6 = 0 (1)
x = 3, −2, both admissible (1)

Question 10 · 3 marks

Find two numbers whose sum is 15 and whose product is 56.

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The numbers are the roots of x² − 15x + 56 = 0. Split −15x as −7x − 8x: x(x − 7) − 8(x − 7) = (x − 7)(x − 8) = 0. The numbers are 7 and 8. Check: 7 + 8 = 15, 7 × 8 = 56.

forms x² − 15x + 56 = 0 (1)
factorises to (x − 7)(x − 8) (1)
numbers 7 and 8 (1)

Question 11 · 3 marks

If α and β are the roots of 2x² − 6x + 3 = 0, find α² + β².

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The roots of the equation are the zeroes of 2x² − 6x + 3, so α + β = −b/a = 3 and αβ = c/a = 3/2. α² + β² = (α + β)² − 2αβ = 9 − 3 = 6.

α + β = 3 (1)
αβ = 3/2 (1)
α² + β² = 6 (1)

Class 10 Maths Chapter 4: Quadratic Equations worksheet with answer key