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]}