Solve each of the following quadratic equations:

$x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ [CBSE 2012]

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Solution

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

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