CBSE • Class 10 • Mathematics • Chapter 4 (Quadratic Equations) • Exercise 4.2

Exercise 4.2: Quadratic Equations — NCERT Solutions

Solving quadratic equations using the quadratic formula and completing the square.

Aligned to the latest NCERT 2024-25 edition • 6 questions in this exercise • Free plan, no credit card

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What this exercise covers

Completing the squareQuadratic formulaReal-world quadratic problems

Step-by-step solutions — Exercise 4.2

6 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 4.2 Q1 • 3 marks

Solve 2x2−7x+3=02x^2 - 7x + 3 = 0 using the quadratic formula.
Hint (Socratic — try this first)▾
What are the values of aa, bb, cc, and what does b2−4acb^2 - 4ac evaluate to?
Step-by-step solution▾

Here a=2a = 2, b=−7b = -7, c=3c = 3.

Discriminant: b2−4ac=(−7)2−4(2)(3)=49−24=25b^2 - 4ac = (-7)^2 - 4(2)(3) = 49 - 24 = 25.

Quadratic formula: x=−b±b2−4ac2a=7±254=7±54x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}

So x=7+54=3x = \dfrac{7+5}{4} = 3 or x=7−54=12x = \dfrac{7-5}{4} = \dfrac{1}{2}.

Roots: x=3x = 3 and x=12x = \dfrac{1}{2}.

Common mistake:
Forgetting that b=−7b = -7, so −b=+7-b = +7; some students write −b=−7-b = -7 and get wrong roots.
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Exercise 4.2 Q2 • 3 marks

Solve x2+4x−5=0x^2 + 4x - 5 = 0 by the method of completing the square.
Hint (Socratic — try this first)▾
What number must you add and subtract to make x2+4xx^2 + 4x a perfect square?
Step-by-step solution▾

Start with x2+4x−5=0x^2 + 4x - 5 = 0, so x2+4x=5x^2 + 4x = 5.

Half of the coefficient of xx is 42=2\frac{4}{2} = 2, and 22=42^2 = 4. Add 44 to both sides: x2+4x+4=5+4x^2 + 4x + 4 = 5 + 4 (x+2)2=9(x + 2)^2 = 9

Taking square roots: x+2=±3x + 2 = \pm 3 x=−2+3=1orx=−2−3=−5x = -2 + 3 = 1 \quad \text{or} \quad x = -2 - 3 = -5

Roots: x=1x = 1 and x=−5x = -5.

Common mistake:
Adding the square term to only one side of the equation, which changes the balance of the equation.
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Exercise 4.2 Q3 • 4 marks

Solve 3x2−5x+2=03x^2 - 5x + 2 = 0 by completing the square.
Hint (Socratic — try this first)▾
Should you divide throughout by the coefficient of x2x^2 first?
Step-by-step solution▾

Divide throughout by 33: x2−53x+23=0⇒x2−53x=−23x^2 - \frac{5}{3}x + \frac{2}{3} = 0 \Rightarrow x^2 - \frac{5}{3}x = -\frac{2}{3}

Half of 53\frac{5}{3} is 56\frac{5}{6}, and (56)2=2536\left(\frac{5}{6}\right)^2 = \frac{25}{36}. Add to both sides: x2−53x+2536=−23+2536x^2 - \frac{5}{3}x + \frac{25}{36} = -\frac{2}{3} + \frac{25}{36} (x−56)2=−24+2536=136\left(x - \frac{5}{6}\right)^2 = \frac{-24 + 25}{36} = \frac{1}{36}

Taking square roots: x−56=±16x - \frac{5}{6} = \pm \frac{1}{6} x=56+16=1orx=56−16=23x = \frac{5}{6} + \frac{1}{6} = 1 \quad \text{or} \quad x = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}

Roots: x=1x = 1 and x=23x = \dfrac{2}{3}.

Common mistake:
Forgetting to divide the whole equation by 33 first, so the completing-the-square step uses a wrong half-coefficient.
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Exercise 4.2 Q4 • 4 marks

The sum of a number and its reciprocal is 103\frac{10}{3}. Find the number by forming and solving a quadratic equation.
Hint (Socratic — try this first)▾
If the number is xx, how do you write its reciprocal, and what equation does the condition give?
Step-by-step solution▾

Let the number be xx. Then x+1x=103x + \dfrac{1}{x} = \dfrac{10}{3}.

Multiply throughout by 3x3x: 3x2+3=10x3x^2 + 3 = 10x 3x2−10x+3=03x^2 - 10x + 3 = 0

Using the quadratic formula with a=3,b=−10,c=3a=3, b=-10, c=3: b2−4ac=100−36=64b^2 - 4ac = 100 - 36 = 64 x=10±646=10±86x = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} x=3orx=13x = 3 \quad \text{or} \quad x = \frac{1}{3}

The number is 33 (its reciprocal is 13\frac{1}{3}) — both solutions describe the same pair.

Common mistake:
Writing the reciprocal condition incorrectly (e.g. as x+x=103x + x = \frac{10}{3}) or forgetting to multiply through by xx to clear the fraction.
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Exercise 4.2 Q5 • 3 marks

Solve the equation 2x2+x−4=02x^2 + x - 4 = 0 using the quadratic formula, giving the roots correct to two decimal places.
Hint (Socratic — try this first)▾
After computing the discriminant, is it a perfect square, and does that require you to approximate the square root?
Step-by-step solution▾

Here a=2a = 2, b=1b = 1, c=−4c = -4.

Discriminant: b2−4ac=1−4(2)(−4)=1+32=33b^2 - 4ac = 1 - 4(2)(-4) = 1 + 32 = 33.

x=−1±334x = \frac{-1 \pm \sqrt{33}}{4}

Since 33≈5.745\sqrt{33} \approx 5.745: x=−1+5.7454=4.7454≈1.19x = \frac{-1 + 5.745}{4} = \frac{4.745}{4} \approx 1.19 x=−1−5.7454=−6.7454≈−1.69x = \frac{-1 - 5.745}{4} = \frac{-6.745}{4} \approx -1.69

Roots (to two decimals): x≈1.19x \approx 1.19 and x≈−1.69x \approx -1.69.

Common mistake:
Rounding 33\sqrt{33} too early or incorrectly, and forgetting the negative root when only taking the ++ sign.
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Exercise 4.2 Q6 • 4 marks

The product of two consecutive positive integers is 306306. Form a quadratic equation and find the integers.
Hint (Socratic — try this first)▾
If the smaller integer is xx, how do you express the next consecutive integer?
Step-by-step solution▾

Let the two consecutive positive integers be xx and x+1x + 1.

Given: x(x+1)=306x(x+1) = 306, so x2+x−306=0x^2 + x - 306 = 0

Using the quadratic formula with a=1,b=1,c=−306a=1, b=1, c=-306: b2−4ac=1+1224=1225b^2 - 4ac = 1 + 1224 = 1225 x=−1±12252=−1±352x = \frac{-1 \pm \sqrt{1225}}{2} = \frac{-1 \pm 35}{2} x=17orx=−18x = 17 \quad \text{or} \quad x = -18

Since the integers are positive, x=17x = 17. Hence the integers are 1717 and 1818.

Common mistake:
Accepting the negative root x=−18x = -18 even though the problem states the integers are positive, or using xx and x+2x+2 (consecutive even) instead of xx and x+1x+1.
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How to approach Exercise 4.2

  1. Re-read the chapter summary first. Open Quadratic Equations and refresh the key concepts: Standard form, Factorisation, Quadratic formula, Discriminant.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Quadratic Equations

  1. Exercise 4.1 — Standard form, identifying quadratic equations, and solving by factorisation.
  2. Exercise 4.2 — Solving quadratic equations using the quadratic formula and completing the square.
  3. Exercise 4.3 — Nature of roots — using the discriminant to classify roots without solving.

← Exercise 4.1

Standard form, identifying quadratic equations, and solving by factorisation.

Exercise 4.3 →

Nature of roots — using the discriminant to classify roots without solving.

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