Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is $4 x^{2}-4 a^{2} x+\left(a^{4}-b^{4}\right)=0$.

Comparing it with $A x^{2}+B x+C=0$, we get

$A=4, B=-4 a^{2}$ and $C=a^{4}-b^{4}$

$\therefore$ Discriminant, $D=B^{2}-4 A C=\left(-4 a^{2}\right)^{2}-4 \times 4 \times\left(a^{4}-b^{4}\right)=16 a^{4}-16 a^{4}+16 b^{4}=16 b^{4}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16 b^{4}}=4 b^{2}$

$\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-\left(-4 a^{2}\right)+4 b^{2}}{2 \times 4}=\frac{4\left(a^{2}+b^{2}\right)}{8}=\frac{a^{2}+b^{2}}{2}$

$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-\left(-4 a^{2}\right)-4 b^{2}}{2 \times 4}=\frac{4\left(a^{2}-b^{2}\right)}{8}=\frac{a^{2}-b^{2}}{2}$

Hence, $\frac{1}{2}\left(a^{2}+b^{2}\right)$ and $\frac{1}{2}\left(a^{2}-b^{2}\right)$ are the roots of the given equation.