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?
 A: $$x+1+x-1-2\sqrt{x+1}\sqrt{x-1}=4x-1\implies(2x-1)^2=4(x^2-1)\implies$$
$$4x^2-4x+1=4x^2-4\implies 4x=5\implies x=\frac54$$
But, indeed
$$\sqrt{\frac54+1}-\sqrt{\frac54-1}\stackrel ?=\sqrt{5-1}\iff\frac32-\frac12=2$$
which is false, thus no solution.
A: 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.
A: 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 $(1),(2)$ 
A: Simply square both sides twice to remove radicals
$$
\begin{align}
2x-2\sqrt{x^2-1}&=4x-1\\
1-2x&=2\sqrt{x^2-1}\\
4x^2-4x+1&=4x^2-4\\
5&=4x\end{align}
$$
Plugging $x=\frac54$ into the original equation yields $1=2$. Thus, there is no solution
