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

Mar 29, 2018

$- 2 < x < 0$

#### Explanation:

Leading degree is 2 which tells us that ${x}^{2} + 2 x$ 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 ${x}^{2} + 2 x = 0$

$x \left(x + 2\right) = 0$

$x = 0 \setminus \mathmr{and} x + 2 = 0$

$x = 0 \setminus \mathmr{and} x = - 2$

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 ${x}^{2} + 2 x < 0$ at $- 2 < x < 0$.