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NCERT Solutions · Class 10 · Chapter 2: Polynomials

NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

Exercise 2.2: Zeroes and coefficients of a quadratic. For \(ax^2 + bx + c\) with zeroes \(\alpha, \beta\): \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\). Conversely, \(x^2 - (\alpha + \beta)x + \alpha\beta\) has zeroes \(\alpha, \beta\).

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Exercise 2.2 questions and solutions

Exercise 2.2, Question 1

Find the zeroes of each quadratic polynomial and verify the relationship between the zeroes and the coefficients.
(i) \(x^2 - 2x - 8\)
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  1. Factorise: \((x - 4)(x + 2)\).
  2. Zeroes: \(4, -2\).
  3. Check: sum \(= 2 = -\dfrac{-2}{1}\) and product \(= -8 = \dfrac{-8}{1}\), matching \(-\dfrac{b}{a}\) and \(\dfrac{c}{a}\).
Answer: Zeroes \(4, -2\); relationships verified
(ii) \(4s^2 - 4s + 1\)
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  1. Factorise: \((2s - 1)^2\).
  2. Zeroes: \(\tfrac12, \tfrac12\).
  3. Check: sum \(= 1 = -\dfrac{-4}{4}\) and product \(= \tfrac14 = \dfrac{1}{4}\), matching \(-\dfrac{b}{a}\) and \(\dfrac{c}{a}\).
Answer: Zeroes \(\tfrac12, \tfrac12\); relationships verified
(iii) \(6x^2 - 3 - 7x\)
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  1. Factorise: \(6x^2 - 7x - 3 = (2x - 3)(3x + 1)\).
  2. Zeroes: \(\tfrac32, -\tfrac13\).
  3. Check: sum \(= \tfrac76 = -\dfrac{-7}{6}\) and product \(= -\tfrac12 = \dfrac{-3}{6}\), matching \(-\dfrac{b}{a}\) and \(\dfrac{c}{a}\).
Answer: Zeroes \(\tfrac32, -\tfrac13\); relationships verified
(iv) \(4u^2 + 8u\)
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  1. Factorise: \(4u(u + 2)\).
  2. Zeroes: \(0, -2\).
  3. Check: sum \(= -2 = -\dfrac{8}{4}\) and product \(= 0 = \dfrac{0}{4}\), matching \(-\dfrac{b}{a}\) and \(\dfrac{c}{a}\).
Answer: Zeroes \(0, -2\); relationships verified
(v) \(t^2 - 15\)
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  1. Factorise: \((t - \sqrt{15})(t + \sqrt{15})\).
  2. Zeroes: \(\sqrt{15}, -\sqrt{15}\).
  3. Check: sum \(= 0 = -\dfrac{0}{1}\) and product \(= -15 = \dfrac{-15}{1}\), matching \(-\dfrac{b}{a}\) and \(\dfrac{c}{a}\).
Answer: Zeroes \(\sqrt{15}, -\sqrt{15}\); relationships verified
(vi) \(3x^2 - x - 4\)
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  1. Factorise: \((3x - 4)(x + 1)\).
  2. Zeroes: \(\tfrac43, -1\).
  3. Check: sum \(= \tfrac13 = -\dfrac{-1}{3}\) and product \(= -\tfrac43 = \dfrac{-4}{3}\), matching \(-\dfrac{b}{a}\) and \(\dfrac{c}{a}\).
Answer: Zeroes \(\tfrac43, -1\); relationships verified

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Exercise 2.2, Question 2

Find a quadratic polynomial whose zeroes have the given sum and product.
(i) Sum of zeroes \(\tfrac14\), product \(-1\)
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  1. A quadratic with zeroes of sum \(S\) and product \(P\) is \(k(x^2 - Sx + P)\).
  2. \(x^2 - (\tfrac14)x + (-1)\); multiply by a convenient \(k\) to clear fractions or surds.
Answer: \(4x^2 - x - 4\)
(ii) Sum of zeroes \(\sqrt2\), product \(\tfrac13\)
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  1. A quadratic with zeroes of sum \(S\) and product \(P\) is \(k(x^2 - Sx + P)\).
  2. \(x^2 - (\sqrt2)x + (\tfrac13)\); multiply by a convenient \(k\) to clear fractions or surds.
Answer: \(3x^2 - 3\sqrt2\,x + 1\)
(iii) Sum of zeroes \(0\), product \(\sqrt5\)
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  1. A quadratic with zeroes of sum \(S\) and product \(P\) is \(k(x^2 - Sx + P)\).
  2. \(x^2 - (0)x + (\sqrt5)\); multiply by a convenient \(k\) to clear fractions or surds.
Answer: \(x^2 + \sqrt5\)
(iv) Sum of zeroes \(1\), product \(1\)
Show solution
  1. A quadratic with zeroes of sum \(S\) and product \(P\) is \(k(x^2 - Sx + P)\).
  2. \(x^2 - (1)x + (1)\); multiply by a convenient \(k\) to clear fractions or surds.
Answer: \(x^2 - x + 1\)
(v) Sum of zeroes \(-\tfrac14\), product \(\tfrac14\)
Show solution
  1. A quadratic with zeroes of sum \(S\) and product \(P\) is \(k(x^2 - Sx + P)\).
  2. \(x^2 - (-\tfrac14)x + (\tfrac14)\); multiply by a convenient \(k\) to clear fractions or surds.
Answer: \(4x^2 + x + 1\)
(vi) Sum of zeroes \(4\), product \(1\)
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  1. A quadratic with zeroes of sum \(S\) and product \(P\) is \(k(x^2 - Sx + P)\).
  2. \(x^2 - (4)x + (1)\); multiply by a convenient \(k\) to clear fractions or surds.
Answer: \(x^2 - 4x + 1\)

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