Answer
no solution
Work Step by Step
Squaring both sides of the given equation, $
\sqrt{x+1}-\sqrt{x-1}=2
,$ results to
\begin{array}{l}\require{cancel}
(\sqrt{x+1}-\sqrt{x-1})^2=(2)^2
\\
(\sqrt{x+1})^2-2(\sqrt{x+1})(\sqrt{x-1})+(\sqrt{x-1})^2=4
\\
x+1-2\sqrt{x+1}\sqrt{x-1}+x-1=4
\\
(x+x)+(1-1-4)=2\sqrt{x+1}\sqrt{x-1}
\\
2x+4=2\sqrt{x+1}\sqrt{x-1}
\\
\dfrac{2x+4}{2}=\dfrac{2\sqrt{x+1}\sqrt{x-1}}{2}
\\
x+2=\sqrt{x+1}\sqrt{x-1}
.\end{array}
Squaring both sides for the second time results to
\begin{array}{l}\require{cancel}
(x+2)^2=(\sqrt{x+1}\sqrt{x-1})^2
\\
(x)^2+2(x)(2)+(2)^2=(x+1)(x-1)
\\
x^2+4x+4=x^2-1
\\
(x^2-x^2)+4x=-1-4
\\
4x=-5
\\
x=-\dfrac{5}{4}
.\end{array}
If $x=-\dfrac{5}{4},$ then the part of the original equation, $-\sqrt{x-1},$ results to a negative radicand. This is not defined for the set of real numbers. Hence, there is no solution.
