Answer
$x=-1$
Work Step by Step
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 $x$:
$x+1=\sqrt{0}$
$x+1=0$
$x=-1$
