We write, $-x=-\frac{x}{2}-\frac{x}{2}$ as $2 x^{2} \times \frac{1}{8}=\frac{x^{2}}{4}=\left(-\frac{x}{2}\right) \times\left(-\frac{x}{2}\right)$

$\begin{array}{l}
\therefore 2 x^{2}-x+\frac{1}{8}=0 \\
\Rightarrow 2 x^{2}-\frac{x}{2}-\frac{x}{2}+\frac{1}{8}=0 \\
\Rightarrow 2 x\left(x-\frac{1}{4}\right)-\frac{1}{2}\left(x-\frac{1}{4}\right)=0 \\
\Rightarrow\left(x-\frac{1}{4}\right)\left(2 x-\frac{1}{2}\right)=0 \\
\Rightarrow x-\frac{1}{4}=0 \text { or } 2 x-\frac{1}{2}=0 \\
\Rightarrow x=\frac{1}{4} \text { or } x=\frac{1}{4}
\end{array}$

Hence, $\frac{1}{4}$ is the repeated root of the given equation.