The equation

$\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1},(x\in \mathbf{R})$

 

The given equation is valid if $x + 1 \ge 0, x – 1$

≥ 0 and 4x – 1 ≥ 0 i.e. if x ≥ 1.

 Squaring both the sides we get

$\begin{array}{l} x+1+x-1-2\sqrt{(x+1)(x-1)}=4x-1\\ \Rightarrow 1-2x=2\sqrt{(x+1)(x-1)}\end{array}$

Squaring again, we get

$\begin{array}{l} 1-4x+4x^2=4\left(x^2-1\right)\\ \Rightarrow 4x=5\text{ or }x=5/4.\end{array}$

Putting this value of x in the given equation, we get

$\begin{array}{l} \sqrt{\frac{5}{4}}+1-\sqrt{\frac{5}{4}-1}=\sqrt{4\left(\frac{5}{4}\right)-1}\\ \Rightarrow \frac{3}{2}-\frac{1}{2}=2\text{ or }1=2\end{array}$

which is not true.

Thus, the given equation has no solution.