# solving inequalities [duplicate]

Possible Duplicate:
Solving a quadratic inequality

Suppose we have $(x-5)(x+2) > 0$. Can we solve it like this: $(x-5) > 0$ and $(x+2) >0$, $x > 5$ and $x > -2$? Please help.

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This is a dupe. –  Tom Stephens Aug 15 '10 at 19:45
I appreciate your interest in Mathematics but at the same time, please make sure that your question is of interest to the users here. If you want to just solve your problems i shall better not recommend this site. Please visit Art of Problem Solving. You shall be much better there. –  anonymous Aug 15 '10 at 20:08
This should be deleted as well –  Casebash Aug 15 '10 at 21:16

## marked as duplicate by Chandru1, Larry WangAug 15 '10 at 20:09

If $ab>0$ does it always imply that $a>0$ or $b>0$. You can also have both $a$ and $b$ to be less than $0$. So in your question it can also be $(x-5)<0$ along with $(x+2)<0$.
@Zia: I feel that this is not the right place to ask such questions. Anyhow i am explaining this. Take 2 real numbers $a,b$ and consider their product $ab$. When can $ab>0$. Either both $a$ and $b$ should be positive or else both $a$ and $b$ should be negative. This is precisely what i have done, which you missed while you posted the question. –  anonymous Aug 15 '10 at 20:01