Class 10 Mathematics Chapter 2 Polynomials Solutions

Exercise 2.1

1. Number of zeroes from graphs:

(i)
Graph does not touch the x-axis.
Number of zeroes = 0

(ii)
Graph touches x‑axis at 1 point.
Number of zeroes = 1

(iii)
Graph cuts x‑axis at 2 points.
Number of zeroes = 2

(iv)
Graph touches x‑axis at 1 point (double root).
Number of zeroes = 1

(v)
Graph cuts x‑axis at 3 points.
Number of zeroes = 3

(vi)
Graph cuts x‑axis at 4 points.
Number of zeroes = 4


Exercise 2.2

Q1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:

(i) For \(x^2 - 2x - 8 = 0\):

\[ x^2 - 2x - 8 = 0 \] \[ x^2 - 4x + 2x - 8 = 0 \] \[ (x - 4)(x + 2) = 0 \] \[ x = 4, -2 \] \[ \text{Sum} = 4 + (-2) = 2 = -\frac{-2}{1}, \quad \text{Product} = 4 \times (-2) = -8 = \frac{-8}{1} \]

(ii) For \(4s^2 - 4s + 1 = 0\):

\[ 4s^2 - 4s + 1 = 0 \] \[ (2s - 1)^2 = 0 \] \[ s = \tfrac12, \tfrac12 \] \[ \text{Sum} = 1, \quad \text{Product} = \tfrac14 \] \[ 1 = -\frac{-4}{4}, \quad \frac14 = \frac{1}{4} \]

(iii) For \(6x^2 - 7x - 3 = 0\):

\[ 6x^2 - 7x - 3 = 0 \] \[ 6x^2 + 2x - 9x - 3 = 0 \] \[ 2x(3x + 1) - 3(3x + 1) = 0 \] \[ (3x + 1)(2x - 3) = 0 \] \[ x = -\tfrac13, \tfrac32 \] \[ \text{Sum} = \tfrac{7}{6} = -\frac{-7}{6}, \quad \text{Product} = -\tfrac12 = \frac{-3}{6} \]

(iv) For \(4u^2 + 8u + 4 = 0\):

\[ 4u^2 + 8u + 4 = 0 \] \[ 4(u^2 + 2u + 1) = 0 \] \[ (u + 1)^2 = 0 \] \[ u = -1, -1 \] \[ \text{Sum} = -2 = -\frac{8}{4}, \quad \text{Product} = 1 = \frac{4}{4} \]

(v) For \(t^2 - 15 = 0\):

\[ t^2 - 15 = 0 \] \[ t = \pm \sqrt{15} \] \[ \text{Sum} = 0 = -\frac{0}{1}, \quad \text{Product} = -15 = \frac{-15}{1} \]

(vi) For \(3x^2 - x - 4 = 0\):

\[ 3x^2 - x - 4 = 0 \] \[ 3x^2 + 2x - 3x - 4 = 0 \] \[ x(3x + 2) - 2(3x + 2) = 0 \] \[ (3x + 2)(x - 2) = 0 \] \[ x = -\tfrac{2}{3}, 2 \] \[ \text{Sum} = \tfrac{4}{3} = -\frac{-1}{3}, \quad \text{Product} = -\tfrac{4}{3} = \frac{-4}{3} \]

Q2: Form quadratic polynomials using the given sums and products of zeroes:

(i) For \(\alpha + \beta = \frac14, \alpha\beta = -1\):

\[ x^2 - \frac14 x - 1 = 0 \] \[ \text{Multiplying by }4:\;4x^2 - x - 4 = 0 \]

(ii) For \(\alpha + \beta = \sqrt{2}, \alpha\beta = \frac13\):

\[ x^2 - \sqrt{2}\,x + \frac13 = 0 \] \[ \text{Multiplying by }3:\;3x^2 - 3\sqrt{2}\,x + 1 = 0 \]

(iii) For \(\alpha + \beta = 0, \alpha\beta = \sqrt{5}\):

\[ x^2 + \sqrt{5}\,x = 0 \]

(iv) For \(\alpha + \beta = 1, \alpha\beta = 1\):

\[ x^2 - x + 1 = 0 \]

(v) For \(\alpha + \beta = -\tfrac14, \alpha\beta = \tfrac14\):

\[ x^2 + \frac14x + \frac14 = 0 \] \[ \text{Multiplying by }4:\;4x^2 + x + 1 = 0 \]

(vi) For \(\alpha + \beta = 4, \alpha\beta = 1\):

\[ x^2 - 4x + 1 = 0 \]
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