Solve each of the following quadratic equations:

$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ [CBSE 2015]

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Solution

We write, $6 x = (a + 4) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + 2 a - 8)] = - (a^{2} + 2 a - 8) x^{2} = (a + 4) x \times [- (a - 2) x]$
$∴ x^{2} + 6 x - (a^{2} + 2 a - 8) = 0 \Rightarrow x^{2} + (a + 4) x - (a - 2) x - (a + 4) (a - 2) = 0 \Rightarrow x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0 \Rightarrow [x + (a + 4)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 4) or x = a - 2$
Hence, $- (a + 4)$ and $(a - 2)$ are the roots of the given equation.

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