The question is this:
x^2 - 9
--------- > 0
x
I pretty much have no idea how to do it, maybe something to do with difference of two squares in the numerator?
Any help would be appreciated
If fraction is to be positive than there are two conditions:
1. numerator and denominator are both positive.
2. numerator and denominator are both negative.
$\displaystyle \frac{x^2-9}{x}$$\displaystyle >0$
$\displaystyle x^2-9>0 , x>3 $ condition x is positive, in this case both numerator and denominator are positive.
For denominator to be negative $\displaystyle x<0$ and for numerator to be negative $\displaystyle -3<x<3$
so $\displaystyle -3<x<0$
So the combine answer is $\displaystyle -3<x<0$ and $\displaystyle x>3$