## Intermediate Algebra (6th Edition)

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.