smc

Subtract $4x$ from both sides of the initial equation and solve for $x$: $$\begin{array}{rl}\sqrt{2x}& =1-4x\\ 2x& =1-8x+16x^2\\ 16x^2-10x+1& =0\\ (8x-1)(2x-1)& =0\end{array}$$ It may look like we have two different solutions for $x$, but plugging in $x=\frac{1}{2}$ into the original equation reveals that it is extraneous. Thus, $x=\frac{1}{8}$, which is only consistent with answer choice $\boxed{\text{is fractional}}$.