How do you solve #x^3-x^2<9x-9#?

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Answer 1 · 2016-07-29T06:49:15

#x < -3# or #1 < x < 3#

This inequality factors as:

#x^2(x-1) < 9(x-1)#

If #x=1# then both sides are #0# and the inequality is false.

Case #bb(x > 1)#

#(x-1) > 0# so we can divide both sides by #(x-1)# to get:

#x^2 < 9#

Hence #-3 < x < 3#

So this case gives solutions #1 < x < 3#

Case #bb(x < 1)#

#(x-1) < 0# so we can divide both sides by #(x-1)# and reverse the inequality to get:

#x^2 > 9#

Hence #x < -3# or #x > 3#

So this case gives solutions #x < -3#