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$$\sqrt{x^2-9}$$

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?

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  • $\begingroup$ You need to solve for when $x^2-9\ge0$. So you get $x^2\ge9$. But when you take the square root of $x^2$, what happens?? $\endgroup$ Aug 22, 2013 at 4:10
  • $\begingroup$ Ultimately your answer is right, but the steps are not. The bad $x$ are the $x$ with $x^2\lt 9$, that is, with $-3\lt x\lt 3$. Now you can write down the good $x$ easily. $\endgroup$ Aug 22, 2013 at 4:11

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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?

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  • $\begingroup$ Could you explain why it's a mistake to add 9 and take the square root of both sides? Can you not take the square root of inequalities? $\endgroup$
    – Kot
    Aug 22, 2013 at 4:14
  • $\begingroup$ Well, you can. However, starting with $x^2<9$ and applying the square root to both sides, we get $$\sqrt{x^2}<\sqrt 9,$$ or equivalently, $$|x|<3.$$ The kicker, here, is that $\sqrt{x^2}=x$ for non-negative $x$ only. In general, the best we can say is that $\sqrt{x^2}=|x|.$ $\endgroup$ Aug 22, 2013 at 4:21
  • $\begingroup$ It's also worth noting that what you're calculating is the points that are not in the domain. The points in the domain are those $x$ such that $x^2\ge 9.$ $\endgroup$ Aug 22, 2013 at 4:25
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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 $x\ge 3$.

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  • $\begingroup$ I am confused why you left out $x>-3$ in your fifth bullet point. Why don't you use $x>-3$ since it includes $x>3$? $\endgroup$
    – Kot
    Aug 22, 2013 at 4:29
  • $\begingroup$ @Steven: Because it doesn’t: $0>-3$, but $0\not>3$. If you know that $x>3$, then you know that $x>-3$, so the requirement that $x>-3$ is superfluous. If you know that $x>-3$, you do not know that $x>3$, so the requirement that $x>3$ is not superfluous. $\endgroup$ Aug 22, 2013 at 4:32
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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 $a=3$.

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  • $\begingroup$ I'll do it, Thanks. $\endgroup$ Aug 22, 2013 at 6:14
  • $\begingroup$ @amWhy: God time Amy. :) $\endgroup$
    – Mikasa
    Aug 23, 2013 at 0:37
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Hint: $\sqrt{x^2}=|x|$ This will help you to understand what is happening..

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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.

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  • $\begingroup$ @AndréNicolas thank you. $\endgroup$ Aug 22, 2013 at 4:17

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