The number of solution(s) of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}$ is/are

Given equation is $$ \sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1} $$ On squaring both sides, we get $$ (x+1)+(x-1)-2 \sqrt{x^{2}-1}=4 x-1 $$ $\Rightarrow \quad 2 x-2 \sqrt{x^{2}-1}=4 x-1$ $\Rightarrow \quad-2 \sqrt{x^{2}-1}=2 x-1$ Again, squaring both sides, we get $$ 4\left(x^{2}-1\right)=4 x^{2}+1-4 x $$ $\Rightarrow \quad-4=+1-4 x$ $\Rightarrow \quad 4 x=5$ $\Rightarrow$ $$ x=\frac{5}{4} $$ But, when we put $x=\frac{5}{4}$ in Eq. (i), we get $$ \sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=\sqrt{4 \times \frac{5}{4}-1} $$ $\Rightarrow$ $\sqrt{\frac{9}{4}}-\sqrt{\frac{1}{4}}=\sqrt{5-1} \Rightarrow \frac{3}{2}-\frac{1}{2}=2$ $\Rightarrow 1=2,$ which is not true. Hence, no value of $x$ satisfy the given equation.