Question

Solve each of the following quadratic equations:

$4x^2-4a^{2}x+\left(a^4-b^4\right)=0$ [CBSE 2015]

Answer 1

We write, $-4a^{2}x=-2\left(a^2+b^2\right)x-2\left(a^2-b^2\right)x$ as $4x^2\times \left(a^4-b^4\right)=4\left(a^4-b^4\right)x^2=\left[-2\left(a^2+b^2\right)\right]x\times \left[-2\left(a^2-b^2\right)\right]x$
$$\begin{gathered} \therefore 4x^2-4a^{2}x+\left(a^4-b^4\right)=0 \\ \Rightarrow 4x^2-2\left(a^2+b^2\right)x-2\left(a^2-b^2\right)x+\left(a^2-b^2\right)\left(a^2+b^2\right)=0 \\ \Rightarrow 2x\left[2x-\left(a^2+b^2\right)\right]-\left(a^2-b^2\right)\left[2x-\left(a^2+b^2\right)\right]=0 \\ \Rightarrow \left[2x-\left(a^2+b^2\right)\right]\left[2x-\left(a^2-b^2\right)\right]=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow 2x-\left(a^2+b^2\right)=0\mathrm{or}2x-\left(a^2-b^2\right)=0 \\ \Rightarrow x=\frac{a^2+b^2}{2}\mathrm{or}x=\frac{a^2-b^2}{2} \end{gathered}$$
Hence, $\frac{a^2+b^2}{2}$ and $\frac{a^2-b^2}{2}$ are the roots of the given equation.