Question

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

Answer 1

`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]}