Find the domain of $\sqrt{x^2-9}$ $$\sqrt{x^2-9}$$
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?
 A: You went a bit off track midway. Instead, let's rewrite $x^2-9<0$ as $(x+3)(x-3)<0.$ This will only be true when $x+3$ and $x-3$ are of opposite sign (why?), so since $x-3<x+3$ for all real $x,$ then we need to figure out when $x-3<0$ and $x+3>0$. Can you take it the rest of the way?
A: You need to have $x^2-9\ge 0$, not $x^2-9<0$. This is $(x-3)(x+3)\ge 0$, which is the case when 


*

*one of $x-3$ and $x+3$ is $0$,  

*both $x-3$ and $x+3$ are positive, or  

*both $x-3$ and $x+3$ are negative.


In all other cases $(x-3)(x+3)$ is negative, which is precisely what you don’t want.


*

*The first of these happens when $x=3$ or $x=-3$. 

*The second happens when $x>3$ and $x>-3$; of course if $x>3$, then automatically $x>-3$, so this happens when $x>3$.

*And the third happens when $x<3$ and $x<-3$; this time we notice that if $x<-3$, then automatically $x<3$, so this happens when $x<-3$.
Putting the pieces together, we see that $x^2-9\ge 0$ when $x\le -3$ or $x\ge 3$.
With practice, however, you should come to realize that if $a\ge 0$, then $x^2\ge a^2$ if and only if $|x|>\sqrt{a}$. Here that means that $|x|\ge 3$, which means that $x$ is at least as far away from $0$ as $3$ is, i.e., that $x\le -3$ or $x\ge 3$.
A: 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 $a=3$.
A: Hint:  $\sqrt{x^2}=|x|$  This will help you to understand what is happening..
A: You had a little bit of a mistake.  You were right when you said that $x$ is in the domain when
$$
x^2 \geq 9
$$
This, in turn, means that $x\color{red}\leq-3$ and $x\geq3$.  So, the domain of this function is $(-\infty,-3]\cup[3,\infty)$ as desired.
