# How do you solve x^2/(x^2+x)>=0?

Sep 11, 2016

$- 1 < x \le 0$

#### Explanation:

Calling $y = {x}^{2} + x > 0$ we have

${x}^{2} + x - y > 0$ or

$\frac{- 1 - \sqrt{1 + 4 y}}{2} < x < \frac{- 1 + \sqrt{1 + 4 y}}{2}$

The smallest interval is for $y = 0$ so

$- 1 < x \le 0$