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,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$

Write the equation as
$$
\sqrt{x+1}=\sqrt{x-1}+\sqrt{4x-1}
$$
Then you must have
\begin{cases}
x+1\ge0\\
x-1\ge0\\
4x-1\ge0
\end{cases}
which boils down to $x\ge1$. Now square, you're sure not to add spurious solutions, because both sides represent non negative numbers:
$$
x+1=x-1+4x-1+2\sqrt{(x-1)(4x-1)}
$$
or
$$
-4x+3=2\sqrt{(x-1)(4x-1)}
$$
Now the right hand side is non negative, so also the left hand side must be, which means
$$
-4x+3\ge0
$$
or $x\le 3/4$. With the previous limitation, this has the consequence that no solution can exist.
