How do you solve $x^2+2x>0$ ?

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Answer 1 · 2017-06-29T11:34:07

Solution: $x < -2 or x>0$ . In interval notation:

$(-oo, -2) uu (0 , oo)$ .

$x^2+2x >0 or x(x+2) > 0$ . Critical points are

$x=0 :.x=0 or x+2 =0 :.x =-2$

Sign Change for $x(x+2)$ :

When $x < -2 ; (-)(-) = + :. x(x+2) > 0$

When $-2< x <0 ; (-)(+) = - :. x(x+2) < 0$

When $x >0 ; (+)(+) = + :. x(x+2) > 0$

Solution: $x < -2 or x>0$ .In interval notation:

$(-oo, -2) uu (0 , oo)$ . The graph also confirms.

graph{x^2+2x [-10, 10, -5, 5]}

How do you solve x^2+2x>0x^2+2x>0?