Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 637: 31

$x=3$

$x=\sqrt 2x+3$ Square both sides: $x^2=(\sqrt 2x+3)^2$ $x^2=2x+3$ $x^2-2x-3=0$ $(x-3)(x+1)=0$ $x-3=0$ or $x+1=0$ $x=3$ or $x=-1$ With $x=3$ then $3=\sqrt 2(3)+3$ $3=3$ With $x=-1$ then $-1=\sqrt 2(-1)+3$ $-1\ne1$ Hence, the solution is $x=3$