How do you solve $- x^{2} + x \geq 0$ $-x^2+x>=0$ ?

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Answer 1 · 2016-09-30T05:06:04

$(0, 1)$ $(0,1)$
or
$0 < x < 1$ $0 < x < 1$

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

$- x (x - 1) > 0$ $-x(x-1)>0$

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

Case 1 +,+:

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

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

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

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