## College Algebra (11th Edition)

$x=-1$
We are given: $\sqrt{x+2}+1=\sqrt{2x+6}$ We square both sides: $(\sqrt{x+2}+1)^{2}=(\sqrt{2x+6})^{2}$ $(\sqrt{x+2}+1)(\sqrt{x+2}+1)=(\sqrt{2x+6})(\sqrt{2x+6})$ $x+2+2\sqrt{x+2}+1=2x+6$ $x+3+2\sqrt{x+2}=2x+6$ $2\sqrt{x+2}=2x-x+6-3$ $2\sqrt{x+2}=x+3$ We square both sides again: $(2\sqrt{x+2})^{2}=(x+3)^{2}$ $4\ (x+2)=x^{2}+6x+9$ And distribute $4$: $4x+8=x^{2}+6x+9$ Put all terms on one side of the equation: $0=x^2+6x+9-4x-8 \\0=x^2+2x+1 \\x^{2}+2x+1=0$ Factor the trinomial: $(x+1)^{2}=0$ Equate each unique factor to zero thej solve for $x$: $x+1=\sqrt{0}$ $x+1=0$ $x=-1$