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?

As $(x+1)-(x-1)=2$ and given that $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}\ \ \ \ (1)$
$$\sqrt{x+1}+\sqrt{x-1}=\frac2{\sqrt{4x-1}}\ \ \ \ (2)$$
On addition, $$2\sqrt{x+1}=\sqrt{4x-1}+\frac2{\sqrt{4x-1}}=\frac{4x+1}{\sqrt{4x-1}}$$
$$\implies 2\sqrt{x+1} \sqrt{4x-1}=4x+1 \ \ \ \ (3)$$
Squaring we get $4x=5\iff x=\frac54$ which does not satisfy  $(1)$ (the given equation) and $(2)$ but satisfies $(3)$
In fact, $x=\frac54$ is a root of $\sqrt{x+1}+\sqrt{x-1}=\sqrt{4x-1}$ and is an Extraneous  root of $(1),(2)$ 
