## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x=8$
Squaring both sides, the given equation is equivalent to \begin{array}{l}\require{cancel} \sqrt{2x}=\sqrt{x+1}+1 \\\\ \left(\sqrt{2x}\right)^2=\left(\sqrt{x+1}+1\right)^2 \\\\ 2x=(\sqrt{x+1})^2+2(\sqrt{x+1})(1)+(1)^2 \\\\ 2x=x+1+2\sqrt{x+1}+1 \\\\ 2x-x-1-1=2\sqrt{x+1} \\\\ x-2=2\sqrt{x+1} .\end{array} Squaring both sides again, the equation above is equivalent to \begin{array}{l}\require{cancel} x-2=2\sqrt{x+1} \\\\ (x-2)^2=(2\sqrt{x+1})^2 \\\\ (x)^2-2(x)(2)+(2)^2=4(x+1) \\\\ x^2-4x+4=4x+4 \\\\ x^2-4x-4x+4-4=0 \\\\ x^2-8x=0 .\end{array} Factoring the $GCF=x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2-8x=0 \\\\ x(x-8)=0 .\end{array} Equating each factor to zero and then solving the variable, then the solutions are $x=\{0,8\} .$ If $x=0,$ then \begin{array}{l}\require{cancel} \sqrt{2x}=\sqrt{x+1}+1 \\\\ \sqrt{2(0)}=\sqrt{0+1}+1 \\\\ \sqrt{0}=\sqrt{1}+1 \\\\ 0=1+1 \\\\ 0=2 \text{ (FALSE)} .\end{array} Hence, only $x=8$ satisfies the original equation.