It's Mathematics Olympiad! - 2

By squaring both sides of $\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = 1$ , we obtain:

$2x + 2\sqrt{x^2 - (2x-1)} = 1 \Rightarrow 2x + 2\sqrt{(x-1)^2} = 1 \Rightarrow 2x + 2|x-1| = 1 \Rightarrow x = \frac{1}{2}$

However, this is an extraneous root since $\sqrt{1/2 + \sqrt{2(1/2)-1}} + \sqrt{1/2 - \sqrt{2(1/2)-1}} = 2 \cdot \sqrt{1/2} = \sqrt{2} \neq 1$ . If we repeat the same process for 2 on the RHS:

$2x + 2\sqrt{x^2 - (2x-1)} = 2^2 \Rightarrow 2x + 2\sqrt{(x-1)^2} = 4 \Rightarrow 2x + 2|x-1| = 4 \Rightarrow x = \frac{3}{2}$

which does satisfy $\sqrt{3/2 + \sqrt{2(3/2)-1}} + \sqrt{3/2 - \sqrt{2(3/2)-1}} = 2(3/2) + 2\sqrt{9/4-2} = 3 + 2(1/2) = 3 + 1 = 4$ . Hence, $\boxed{x = \frac{3}{2}}$ is the only permissible non-negative solution.