TITLE: Find the domain of $\sqrt{x^2-9}$
QUESTION [5 upvotes]: $$\sqrt{x^2-9}$$
I know that the domain of square root is greater than or equal to zero. I solve for when $x^2-9<0$ and get $x^2<9$. Now I get $x<-3$ and $x<3$. I know that the domain is $(-\infty,-3] \cup [3,\infty)$ The problem is I do not understand how the $x<3$ gets flipped to $x>3$, am I doing this step properly or is there another way to do it?

REPLY [3 votes]: All other answers made a good survey to this question, but I'd like to note  some good points. When we have a function like $f(x)=\sqrt[n]{g(x)}$ while $n$ is even number so:

$f(x)\geq 0$
$g(x)\geq0$

And you see that @Brian's trying to show you that we have to make $x^2-9$ positive. Moreover a very simple effective rule as follows: 

$$x^2-a^2\geq 0\Longleftrightarrow x^2\geq a^2\Longleftrightarrow x\geq a\cup x\leq -a$$ 

This is what @Omno indicated. Here we have $a=3$.