## Precalculus (6th Edition)

$x=-1$
Square both sides: $(\sqrt{x+2}+1)^2=(\sqrt{2x+6})^2 \\(\sqrt{x+2})^2+2(\sqrt{x+2})(1)+1^2=2x+6 \\x+2+2\sqrt{x+2}+1=2x+6 \\2\sqrt{x+2}+x+3=2x+6$ Subtract $x$ and $3$ to both sides of the equation to obtain: \begin{array}{ccc} &2\sqrt{x+2}+x+3-x-3 &= &2x+6-x-3 \\&2\sqrt{x+2} &= &x+3 \end{array} Square both sides of the equation to obtain: $$\left(2\sqrt{x+2}\right)^2=(x+3)^2 \\4(x+2)=x^2+2(x)(3)+3^2 \\4(x)+4(2)=x^2+6x+9 \\4x+8=x^2+6x+9$$ Subtract $4x$ and $8$ to both sides of the equation to obtain: \begin{array}{ccc} &4x+8-4x-8 &= &x^2+6x+9-4x-8 \\&0 &= &x^2+2x+1 \end{array} Factor the trinomial to obtain: $$0=(x+1)(x+1)$$ Use the Zero-Factor Property by equating each factor to zero, then solve each equation to obtain: $x+1=0$ or $x+1=0$ $x=-1$ or $x=-1$ Thus, the solutions is $x=-1$.