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

The given equation is $2 x^{2}-2 \sqrt{2} x+1=0$.

Comparing it with $a x^{2}+b x+c=0$, we get

$a=2, b=-2 \sqrt{2}$ and $c=1$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-2 \sqrt{2})^{2}-4 \times 2 \times 1=8-8=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{2})+\sqrt{0}}{2 \times 2}=\frac{2 \sqrt{2}}{4}=\frac{\sqrt{2}}{2}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{2})-\sqrt{0}}{2 \times 2}=\frac{2 \sqrt{2}}{4}=\frac{\sqrt{2}}{2}$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.