## Precalculus (6th Edition)

$\color{blue}{x=3}$
Add $\sqrt{2x+3}$ to both sides to obtain: $x=\sqrt{2x+3}$ Square both sides to obtain: $x^2=2x+3$ Move all terms to the left side of the equation. Note that when a term is transferred to the other side of the equation, its sign changes to its opposite. $x^2-2x-3=0$ Factor the trinomial to obtain: $(x-3)(x+1)=0$ Use the zero-factor property by equating each factor to zero. $x-3=0$ or $x+1=0$ Solve each equation to obtain: $x=3$ or $x=-1$. $-1$ cannot be a solution to the equation as $\sqrt{2x+3}$ is never negative. Thus, the solution to the given equation is: $\color{blue}{x=3}$.