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I'm in the last year of high school, but my school's math program is really bad and I'm trying to catch up on my math. I've been scratching my head on this one but I can't seem to solve it :(

For $x^2+2x > 0$, why are the solutions $x < -2$ and $x > 0$?

I tried working it out but I only get $x > -2$ and $x > 0$?

$x(x+2) > 0$

let $x+2 > 0$

$x > -2$

Thank you!

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When passing from $x(x+2) \gt 0$ to $x+2 \gt 0$ you need to be careful: what if $x \lt 0$? Treat the cases $x \gt 0$ and $x \lt 0$ separately. In the latter case you'll find the solution you missed. – t.b. Apr 14 '12 at 9:32
up vote 3 down vote accepted

Factor $x^2+2x$ as $x(x+2)$. Now consider the possibilities for the signs of the factors $x$ and $x+2$:

     x:       -        -           -          0          +  
   x+2:       -        0           +          +          +  
                      -2                      0

In order for $x(x+2)$ to be positive, both factors must have the same sign: either both are positive, or both are negative. You can see from the diagram that both are negative when $x<-2$, and both are positive when $x>0$, so $x^2+2x>0$ when $x<-2$ OR $x>0$.

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$x(x+2)>0$ if and only if $x>0$ and $x+2>0$ or $x<0$ and $x+2<0$, that is $x>0$ and $x>-2$ or $x<0$ and $x<-2$, so $x>0$ or $x<-2$. Graph like this would help you.

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