## College Algebra (10th Edition)

$x=4$
Square both sides: $(\sqrt{2x+1})^2=3^2 \\2x+1=9$ Subtract $1$ to both sides of the equation: $2x+1-1=9-1 \\2x=8$ Divide $2$ to both sides of the equation: $\frac{2x}{2} = \frac{8}{2} \\x=4$ Check if the solution is valid: $\sqrt{2\cdot4 + 1} = 3 \\\sqrt{9}=3 \\3=3$ True. Thus, the solution is $x=4$.