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,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$
 A: Whenever we square,  we immediately introduce extraneous root
Observe that
$\displaystyle\frac54$   is actually a root of $$\sqrt{x+1}=\sqrt{4x-1}-\sqrt{x-1}$$
Also, observe that $\displaystyle2x-1=-2\sqrt{x^2-1}\le0\implies 2x\le1\iff x\le\frac12$ for real $x$
But, $\displaystyle{\sqrt{x-1}}$ is not real unless $x\ge1$
A: The process of solving an equation is basically that of inversion: you successively apply (inverse) functions to both sides of the equation until you reach a point where the solution is clear.  This process depends on each successive equation (upon applying various inverses successively) being equivalent to the previous one so that the final equation $x=\ldots$ is equivalent to the original equation.  However, when you apply non-invertible operations (such as $x\mapsto x^{2}$, i.e. squaring both sides), you don't get an equivalence between the equation before squaring and the equation after squaring: you get a forward implication, which is to say that the final equation $x=\ldots$ does not imply the previous equation(s) prior to squaring, it is only implied itself by the previous chain of equations.
A: 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.
A: 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.
A: $-\frac{x}{x}= \frac{x}{x}$ 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.
A: An equation $F(x)=0$ is an implicit way of defining a set $S$. What we want is an explicit presentation of $S$, say, a list $S=\{a,b,c\}$.
Algebraic manipulations of the form
$$F(x)=0\quad\Rightarrow \quad G(x)=0\quad\Rightarrow\quad\ldots\quad\Rightarrow \quad x\in S'\ ,\tag{1}$$
where $S'$ is a certain finite list (or similar object) do not prove that $S=S'$. They only prove that each $x\in S$ also is in $S'$, in other words: Such manipulations only prove $S\subset S'$. When $S'$ is  a finite set it is usually simple to decide which $x\in S'$ are actually solutions of the original problem.
When the arrows in $(1)$ are reversible, i.e. can be replaced by $\Leftrightarrow$'s with no harm done, then we of course have $S'=S$, and no extra verification is necessary. 
Another instance where $S'$ is automatically the correct solution set is the following: The given equation $F(x)=0$ is of a type for which we have a general theory guaranteeing "exactly two solutions" (as in the case of a quadratic equation) or "exactly one solution", or "a solution space of dimension $d$". When the $S'$ found in $(1)$ satisfies the requirements promised by the general theory then automatically $S'=S$.
A: $\newcommand{\+}{^{\dagger}}
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Indeed, you are solving $\large 4!!!$ equations due to the 'squaring' in your procedure:
$$
\root{x + 1} \pm \root{x - 1} = \pm\root{4x - 1}
$$
For example, $x = 5/4$ satisfies $\root{x + 1} + \root{x - 1} = \root{4x - 1}$
since
$$
\root{{5 \over 4} + 1} \color{#f00}{\Large +} \root{{5 \over 4} - 1}
= {3 \over 2} + \half= 2= \color{#f00}{\Large +}\root{4\times{5 \over 4} - 1}
$$
