Short-Answer Questions

Solve: $4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ [CBSE 2015]

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Solution

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 4 b x - (a - b) (a + b) = 0 \Rightarrow 4 x^{2} + 2 [(a + b) - (a - b)] x - (a - b) (a + b) = 0 \Rightarrow 4 x^{2} + 2 (a + b) x - 2 (a - b) x - (a - b) (a + b) = 0$
$\Rightarrow 2 x [2 x + (a + b)] - (a - b) [2 x + (a + b)] = 0 \Rightarrow [2 x + (a + b)] [2 x - (a - b)] = 0 \Rightarrow 2 x + (a + b) = 0 or 2 x - (a - b) = 0 \Rightarrow x = - \frac{a + b}{2} or x = \frac{a - b}{2}$

Hence, $- \frac{a + b}{2}$ and $\frac{a - b}{2}$ are the roots of the given equation.

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