An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening?

Question

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

Answer 1

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.

Answer 2

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 $x\ge1$

Answer 3

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.

Answer 4

$-\frac{x}{x}= \frac{x}{x}$ 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.

Answer 5

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.

Answer 6

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 $S'=S$.