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Nature of Roots

x 2-2 a x+a 2...

Question

$x^{2} - 2 a x + (a^{2} - b^{2}) = 0$

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Solution

$Given: x^{2} - 2 a x + (a^{2} - b^{2}) = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 1, B = - 2 a and C = (a^{2} - b^{2}) Discriminant D is given by : D = B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times (a^{2} - b^{2}) = 4 a^{2} - 4 a^{2} + 4 b^{2} = 4 b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 a) + \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a + 2 b}{2} = \frac{2 (a + b)}{2} = (a + b) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 a) - \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a - 2 b}{2} = \frac{2 (a - b)}{2} = (a - b) Hence, the roots of the equation are (a + b) and (a - b) .$

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