About the equation $ax^2 - 2x\sqrt {2} + c = 0$, with $a$ and $c$ real constants, 
we are told that the discriminant is zero. The roots are necessarily: 
$\textbf{(A)}\ \text{equal and integral}\qquad \textbf{(B)}\ \text{equal and rational}\qquad \textbf{(C)}\ \text{equal and real} \\ \textbf{(D)}\ \text{equal and irrational} \qquad \textbf{(E)}\ \text{equal and imaginary}$
Plugging into the quadratic formula, we get
\[x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.\]
The discriminant is equal to 0, so this simplifies to $x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.$ Because we are given that $a$ is real, $x$ is always real, and the answer is $\boxed{\textbf{(C)}}.$
~ cxsmi (significant edits)