# How do you solve -x^2+x>=0?

Sep 30, 2016

$\left(0 , 1\right)$
or
$0 < x < 1$

#### Explanation:

Factor $- {x}^{2} + x > 0$

$- x \left(x - 1\right) > 0$

Since this is greater than 0, then the two cases are +,+ and -,-.

Case 1 +,+:

$x - 1 > 0$ becomes $x > 1$
$- x > 0$ becomes $x < 0$ the signs are flipped because we multiplied/divided by a negative.

Draw them on a number line and see where they share their domains. This one doesn't have any domain in common so the inequality doesn't work.

Case 2 -,-:

$x - 1 < 0$ becomes $x < 1$
$- x < 0$ becomes $x > 0$

If you draw this one on a number line, they share the domain at (0,1)
which is when 0 < x < 1.