Question

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

Answer 1

`-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`.