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

1 Answer
Sep 30, 2016

Answer:

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

Explanation:

Factor #-x^2 +x >0#

#-x(x-1)>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.