The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we take $f(x)=\sqrt{(x+1)} -\sqrt{(x-1)} -\sqrt{(4x-1)}$ then we can see that
$f(x)$ is defined when $x \geq 1$ and $f(1) \geq 0\mbox{ and }f'(x) \geq 0$ so, the function is monotone increasing and it will never appear zero.
so, my question is , In this type of equation where the solution actually does not exist, then why should we get this type of solution?
my solution procedure is,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$
Best Answer
Let's consider a more simple example, to understand. Given the equation $$ x = 1 $$ you can take the square of both sides: $$ x^2 = 1 $$ and find two solutions: $$ x=1 \qquad x=-1. $$
This happens because the operation $x\mapsto x^2$ is not invertible. If you apply a non invertible function to an equation, the number of solutions might increase.