Question

If $x$ is real, then find the solution of $\sqrt{x+1}+\sqrt{x-1}=1$

• A. $\frac{5}{4}$
• B. $\frac{4}{5}$
• C. $\frac{3}{5}$
• D. No such $x$ exists.

Answer 1

The correct option is D No such $x$ exists.

On checking the domain we can see that $x>1$

On squaring both sides we have,$2x+2\sqrt{(x^2-1)}=1$

$2\sqrt{(x^2-1)}=1-2x$

On squaring both sides we get, $4(x^2-1)=1+4x^2-4x$.

On solving we get $x=\frac{5}{4}$.

But whenever we are squaring an equation on both sides we get some additional roots which were not meant to be present in the original equation.

So in such case lets substitute $x=\frac{5}{4}$ we get LHS=2 but RHS=1

So $x=\frac{5}{4}$ is the additional root and no such real $x$ exists.

$\therefore$ We get no solution for $x$.