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

The given equation is $x^{2}-4 a x-b^{2}+4 a^{2}=0$.

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

$A=1, B=-4 a$ and $C=-b^{2}+4 a^{2}$

$\therefore$ Discriminant, $D=B^{2}-4 A C=(-4 a)^{2}-4 \times 1 \times\left(-b^{2}+4 a^{2}\right)=16 a^{2}+4 b^{2}-16 a^{2}=4 b^{2}>0$

So, the given equation has real roots.

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

$\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-(-4 a)+2 b}{2 \times 1}=\frac{4 a+2 b}{2}=2 a+b$

$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-(-4 a)-2 b}{2 \times 1}=\frac{4 a-2 b}{2}=2 a-b$

Hence, $(2 a+b)$ and $(2 a-b)$ are the roots of the given equation.