## College Algebra 7th Edition

$x=4$
$\sqrt{2x+1}+1=x$ $\sqrt{2x+1}=x-1$ Square the both sides $\Rightarrow(\sqrt{2x+1})^2=(x-1)^2$ Expand $\Rightarrow2x+1=x^2-2x+1$ $\Rightarrow x^2-4x=0$ Factor $x(x-4)=0$ $x=0$ or $x-4=0\Rightarrow x=4$ Check the answer: $x=0$ LHS $\sqrt{2(0)+1}+1=\sqrt{1}+1=1+1=2$ RHS $0$ LHS $\ne$ RHS $x=0 \color{red}{reject}$ $x=4$ LHS $\sqrt{2(4)+1}+1=\sqrt{8+1}+1=\sqrt{9}+1=3+1=4$ RHS $4$ LHS $=$ RHS $x=4$