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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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marked as duplicate by Chandru1, Larry Wang Aug 15 '10 at 20:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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
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1 Answer

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 $(x+2)<0$.

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i could not understan it what you have written please explain –  Zia ur Rahman Aug 15 '10 at 19:56
    
@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
    
@Zia: What was wrong with the answers provided to your question when it was asked the first time? –  Tom Stephens Aug 15 '10 at 20:01
    
@Tom Stephens: Oh! he is asking the same problem again. Its my mistake of answering. –  anonymous Aug 15 '10 at 20:03
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@All: I request everyone to close, this question as Tom Pointed out this is an exact duplicate. –  anonymous Aug 15 '10 at 20:07
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