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

Feb 17, 2017

$x \in \left(- 2 , 0\right)$

#### Explanation:

Re-write as:

$x \left(x + 2\right) < 0$
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.

$x < 0 \mathmr{and} x + 2 > 0$ demands that both $x > - 2 \mathmr{and} x < 0$ be true. A solution is therefore $x \in \left(- 2 , 0\right)$

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

Here, $x > 0 \mathmr{and} x + 2 < 0$ demands that both $x < - 2 \mathmr{and} x > 0$ be true, which is impossible.

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