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

1 Answer
Feb 17, 2017

Answer:

#x in (-2, 0)#

Explanation:

Re-write as:

#x(x+2) < 0#
We want the interval(s) over which the LHS is less than 0, ie negative.

This will be when either:

  • #x# is negative and #x+2# is positive.

#x< 0 and x+2 >0# demands that both #x> -2 and x< 0# be true. A solution is therefore #x in (-2, 0)#

  • #x# is positive and #x+2# is negative.

Here, #x> 0 and x+2 <0# demands that both #x<-2 and x> 0# be true, which is impossible.

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