Answer
$x=-2$
Work Step by Step
Subtract $2$ from both sides:
$\sqrt{x+6}+2-2=x+6-2$
$\sqrt{x+6}=x+4$
Square both sides:
$(\sqrt{x+6})^2=(x+4)^2$
$x+6=(x+4)^2$
You can square the binomial using the FOIL method:
$(x+4)^2=(x+4)(x+4)$
First: $x\cdot x=x^2$
Outer: $x\cdot 4=4x$
Inner: $x\cdot 4=4x$
Last: $4\cdot 4=16$
Combine them all to get: $x^2+8x+16$
Thus, the equation above becomes:
$x+6=x^2+8x+16$
Subtract $x$ from both sides:
$x-x+6=x^2+8x-x+16$
$6=x^2+7x+16$
Subtract $6$ from both sides:
$6-6=x^2+7x+16-6$
$0=x^2+7x+10$
Factor:
$0=x^2+7x+10$
The leading term of the quadratic trinomial is $x^2$, so you can factor it by finding two numbers that when multiplied together that equal $10$, and added together equal $7$. Those two numbers are $5$ and $2$:
Hence,
$x^2+7x+10=(x+5)(x+2)$
Thus, the equation above is euqivalent to:
$0=(x+5)(x+2)$
Use the Zero-Product Property by equating each factor to zero, then solve each equation to obtain:
\begin{align*}
x+5&=0 &\text{or}& &x+2=0\\
x&=-5 &\text{or}& &x=-2
\end{align*}
Substitute both answers into original problem to check for extraneous solutions:
For $x=-2:$
$\sqrt{(-2)+6}+2=(-2)+6$
$\sqrt{4}+2=4$
$2+2=4$
$4=4$
$x=-2$ is not an extraneous solution
For $x=-5:$
$\sqrt{(-5)+6}+2=(-5)+6$
$\sqrt{1}+2=1$
$1+2=1$
$3/ne1$
$x=-5$ is an extraneous solution.
Therefore the only solution is $x=-2$.
