Solving a radical equation $\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}$ $$
\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}
$$
How many solutions does it have for $x \in \mathbb{R}$?
I squared it once, then rearranges terms to isolate the radical, then squared again.
I got a linear equation, which yielded $x = \frac54$, but when I put that back in the equation, it did not satisfy.
So I think there is no solution, but my book says there is 1.
Can anyone confirm if there is a solution or not?

Square both sides to obtain $2x-2\sqrt{x+1}\sqrt{x-1}=4x-1$ or $-2\sqrt{x+1}\sqrt{x-1}=2x-1$.
Since the left side is not positive, $2x-1\leq0$ or $ x\leq \dfrac12$. But $x\geq 1$, so there is no real solution.
