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NCERT Solutions · Class 10 · Chapter 4: Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

Exercise 4.2: Solving by factorisation. Split the middle term: find two numbers whose product is \(ac\) and whose sum is \(b\), factorise, and set each factor to zero. In word problems, reject roots that make no sense (negative lengths, ages or counts) and say why.

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Exercise 4.2 questions and solutions

Exercise 4.2, Question 1

Find the roots by factorisation.
(i) \(x^2 - 3x - 10 = 0\)
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  1. Two numbers with product \(-10\) and sum \(-3\): \(-5\) and \(2\).
  2. \[\begin{aligned}x^2 - 5x + 2x - 10 &= x(x - 5) + 2(x - 5) \\ &= (x - 5)(x + 2) \\ &= 0\end{aligned}\]
Answer: \(x = 5\) or \(x = -2\)
(ii) \(2x^2 + x - 6 = 0\)
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  1. \(ac = -12\); two numbers with product \(-12\) and sum \(1\): \(4\) and \(-3\).
  2. \[\begin{aligned}2x^2 + 4x - 3x - 6 &= 2x(x + 2) - 3(x + 2) \\ &= (2x - 3)(x + 2) \\ &= 0\end{aligned}\]
Answer: \(x = \dfrac{3}{2}\) or \(x = -2\)
(iii) \(\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0\)
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  1. \(ac = \sqrt{2} \times 5\sqrt{2} = 10\); two numbers with product \(10\) and sum \(7\): \(5\) and \(2\).
  2. Split the middle term: \(\sqrt{2}\,x^2 + 5x + 2x + 5\sqrt{2} = 0\).
  3. Group in pairs: \[x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) = 0\]
  4. Factorise: \((\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0\).
Answer: \[\begin{aligned}x &= -\dfrac{5}{\sqrt{2}} \\ &= -\dfrac{5\sqrt{2}}{2}\end{aligned}\] or \(x = -\sqrt{2}\)

Where marks slip: Write \(2x\) as \(\sqrt{2} \times \sqrt{2}\,x\) to see the common factor \(\sqrt{2}\).

(iv) \(2x^2 - x + \dfrac{1}{8} = 0\)
Show solution
  1. Multiply through by \(8\) to clear the fraction: \(16x^2 - 8x + 1 = 0\).
  2. This is a perfect square: \((4x - 1)^2 = 0\).
Answer: \(x = \dfrac{1}{4}\) (a repeated root: \(\dfrac14, \dfrac14\))
(v) \(100x^2 - 20x + 1 = 0\)
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  1. \[\begin{aligned}100x^2 - 20x + 1 &= (10x)^2 - 2(10x)(1) + 1^2 \\ &= (10x - 1)^2 \\ &= 0\end{aligned}\]
Answer: \(x = \dfrac{1}{10}\) (repeated)

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Exercise 4.2, Question 2

(i) Two friends have 45 marbles between them. Each loses 5, and the product of what they have left is 124. How many did each start with?
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  1. Let one friend start with \(x\); the other has \(45 - x\).
  2. After losing 5 each: \((x - 5)(40 - x) = 124\).
  3. Expand: \[\begin{aligned}&-x^2 + 45x - 200 = 124 \\ \Rightarrow\ &x^2 - 45x + 324 = 0\end{aligned}\]
  4. Product \(324\), sum \(-45\): \(-9\) and \(-36\). So \((x - 9)(x - 36) = 0\).
Answer: One started with \(36\) marbles and the other with \(9\).
(ii) A workshop makes \(x\) toys in a day; each toy costs \(55\) minus the number made that day (in ₹), and the day's total cost is ₹750. Find the number of toys.
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  1. Cost per toy \(= 55 - x\), so total cost \(x(55 - x) = 750\).
  2. \(x^2 - 55x + 750 = 0\). Product \(750\), sum \(-55\): \(-25\) and \(-30\).
  3. \((x - 25)(x - 30) = 0\).
Answer: \(25\) toys or \(30\) toys (both give ₹750).

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Exercise 4.2, Question 3

Find two numbers whose sum is \(27\) and product is \(182\).
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  1. Let the numbers be \(x\) and \(27 - x\): \(x(27 - x) = 182\).
  2. \(x^2 - 27x + 182 = 0\). Product \(182\), sum \(-27\): \(-13\) and \(-14\).
  3. \((x - 13)(x - 14) = 0\), so \(x = 13\) or \(14\); either way the pair is \(13\) and \(14\).
Answer: \(13\) and \(14\)

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Exercise 4.2, Question 4

Find two consecutive positive integers whose squares add up to \(365\).
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  1. Let the integers be \(x\) and \(x + 1\): \(x^2 + (x + 1)^2 = 365\).
  2. \[\begin{aligned}&2x^2 + 2x + 1 = 365 \\ \Rightarrow\ &x^2 + x - 182 = 0\end{aligned}\]
  3. \((x + 14)(x - 13) = 0\). Reject \(x = -14\) (not positive), so \(x = 13\).
Answer: \(13\) and \(14\)

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Exercise 4.2, Question 5

A right triangle has hypotenuse \(13\) cm and its altitude is \(7\) cm less than its base. Find the other two sides.
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  1. Let the base be \(x\) cm; the altitude is \((x - 7)\) cm.
  2. Pythagoras: \[\begin{aligned}&x^2 + (x - 7)^2 = 13^2 \\ \Rightarrow\ &2x^2 - 14x + 49 = 169 \\ \Rightarrow\ &x^2 - 7x - 60 = 0\end{aligned}\]
  3. \((x - 12)(x + 5) = 0\). A length cannot be negative, so \(x = 12\).
Answer: Base \(12\) cm, altitude \(5\) cm

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Exercise 4.2, Question 6

On one day a potter makes some articles; the cost of each is ₹3 more than twice the number made, and the total cost is ₹90. Find the number of articles and the cost of each.
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  1. Let the number be \(x\); cost of each \(= 2x + 3\). Total: \(x(2x + 3) = 90\).
  2. \(2x^2 + 3x - 90 = 0\). \(ac = -180\); product \(-180\), sum \(3\): \(15\) and \(-12\).
  3. \[\begin{aligned}2x^2 + 15x - 12x - 90 &= x(2x + 15) - 6(2x + 15) \\ &= (2x + 15)(x - 6) \\ &= 0\end{aligned}\]
  4. \(x = -\dfrac{15}{2}\) is impossible for a count, so \(x = 6\); each costs \(2(6) + 3 = 15\).
Answer: \(6\) articles at ₹\(15\) each

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Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.