Answer: B

Small Hint:

Set $u=\sqrt{2x},$ so $x=\frac{u^2}{2}$ and $u\ge0$

Big Hint:

Solve the resulting quadratic $2u^2+u-1=0$

Solution:

Let $u=\sqrt{2x}\ge0.$ Since $x=\frac{u^2}{2},$ the equation becomes $$2u^2+u=1,$$ or $(2u-1)(u+1)=0.$ Thus $u=\frac{1}{2}$ and $x=\frac{u^2}{2}=\frac{1}{8}.$ The other quadratic root is inadmissible because $u\ge0.$ Hence $x$ is fractional.

Therefore, the correct answer is B.