An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? 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,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$

$-\frac{x}{x}= \frac{x}{x}$ has no solutions at all, since $-\frac{x}{x}\neq \frac{x}{x}$ no matter what $x$ is.
But we can square both sides, and then what happens?
$\frac{x^2}{x^2}= \frac{x^2}{x^2}$ is an equation that is true for all nonzero numbers. 
By applying a non-invertible operation to both sides, we can turn an equation with no solutions into one with uncountably infinitely many solutions.
