Solve each of the following quadratic equations:

We write, $-4 a^{2} x=-2\left(a^{2}+b^{2}\right) x-2\left(a^{2}-b^{2}\right) x$ as $4 x^{2} \times\left(a^{4}-b^{4}\right)=4\left(a^{4}-b^{4}\right) x^{2}=\left[-2\left(a^{2}+b^{2}\right)\right] x \times\left[-2\left(a^{2}-b^{2}\right)\right] x$

$\therefore 4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$

$\Rightarrow 4 x^{2}-2\left(a^{2}+b^{2}\right) x-2\left(a^{2}-b^{2}\right) x+\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)=0$

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

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

$\Rightarrow 2 x-\left(a^{2}+b^{2}\right)=0$ or $2 x-\left(a^{2}-b^{2}\right)=0$

$\Rightarrow x=\frac{a^{2}+b^{2}}{2}$ 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.