How do you solve #x^2+2x<0#?

1 Answer
Mar 29, 2018

Answer:

#-2< x < 0#

Explanation:

Leading degree is 2 which tells us that #x^2+2x# is quadratic function and leading coefficient is 1 which tells us that parabola opens upward.

Now we can find roots to give us more details.

Let #x^2+2x = 0#

#x(x+2)=0#

#x = 0 \or x+2=0#

#x = 0 \or x=-2#

This tells us that parabola passes x-axis at -2 and then again at 0.

So since parabola opens upward, it would below x-axis at anywhere between #-2# and #0# exclusive, as seen in graph:
graph{ x^2+2x [-3, 1, -2, 1]}Thus #x^2+2x < 0# at #-2 < x < 0#.