Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
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
add comment

2 Answers 2

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 $x>0$.

share|improve this answer
add comment

$x(x+2)>0$ 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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.