2011/12 AHSMC Part 1 Problem 7

Squaring both sides of $2x^2 - 2x - 1 = 2x\sqrt{x^2 - 2x}$, we have $4x^4 - 8x^3 + 4x + 1 = 4x^4 - 8x^3$, which simplifies to $4x+1 = 0$. Hence the only solution is $x = -\frac{1}{4}$. Indeed, $2\bigl(-\frac{1}{4}\bigr)^2 - 2\bigl(-\frac{1}{4}\bigr) = \frac{5}{8}$ and $2\bigl(-\frac{1}{4}\bigr)\sqrt{\bigl(-\frac{1}{4}\bigr)^2 - 2\bigl(-\frac{1}{4}\bigr)} + 1 = \frac{5}{8}$. The answer is (B).