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Question

# Solve each of the following quadratic equations: $4{x}^{2}-4{a}^{2}x+\left({a}^{4}-{b}^{4}\right)=0$ [CBSE 2015]

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Solution

## We write, $-4{a}^{2}x=-2\left({a}^{2}+{b}^{2}\right)x-2\left({a}^{2}-{b}^{2}\right)x$ as $4{x}^{2}×\left({a}^{4}-{b}^{4}\right)=4\left({a}^{4}-{b}^{4}\right){x}^{2}=\left[-2\left({a}^{2}+{b}^{2}\right)\right]x×\left[-2\left({a}^{2}-{b}^{2}\right)\right]x$ $\therefore 4{x}^{2}-4{a}^{2}x+\left({a}^{4}-{b}^{4}\right)=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}-2\left({a}^{2}+{b}^{2}\right)x-2\left({a}^{2}-{b}^{2}\right)x+\left({a}^{2}-{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒2x\left[2x-\left({a}^{2}+{b}^{2}\right)\right]-\left({a}^{2}-{b}^{2}\right)\left[2x-\left({a}^{2}+{b}^{2}\right)\right]=0\phantom{\rule{0ex}{0ex}}⇒\left[2x-\left({a}^{2}+{b}^{2}\right)\right]\left[2x-\left({a}^{2}-{b}^{2}\right)\right]=0$ $⇒2x-\left({a}^{2}+{b}^{2}\right)=0\mathrm{or}2x-\left({a}^{2}-{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{{a}^{2}+{b}^{2}}{2}\mathrm{or}x=\frac{{a}^{2}-{b}^{2}}{2}$ Hence, $\frac{{a}^{2}+{b}^{2}}{2}$ and $\frac{{a}^{2}-{b}^{2}}{2}$ are the roots of the given equation.

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