Question

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

Answer 1

`-1 < x le 0`

Explanation:

Calling `y = x^2+x>0` we have

`x^2+x-y > 0` or

`(-1-sqrt(1+4y))/2 < x < (-1+sqrt(1+4y))/2`

The smallest interval is for `y = 0` so

`-1 < x le 0`