X^3 < X

[Alexander]

Hi -

This is a QC question. I solved this algebra and want to make sure my reasoning is correct.

X^3< X

A: X^2-X
B: 0


1) Subtract and Factor --> X^3 - X < 0
2) Factor --> X ( X^2 - X) < 0
3) The terms to the left of the inequality cannot be 0 because it has to be less than 0 not less than or equal
4) Either X has to be positive and X^2-X be negative or visca versa
5) The answer has to be D because X^2-X can be positive or negative.

Is my thinking correct?

Thanks,
Alexander

[tommywallach]

Hey Alexander,

Hm. You're right that x (x^2 - x) < 0 means that either x is positive and (x^2 - x) is negative or vice-versa, but it's possible that one of those outcomes isn't actually possible. What I mean is that you haven't sorted out whether, when x is negative, x^2 - x is ALWAYS negative. If that were true, then x would HAVE to be positive, which would force (x^2 - x) to be negative, and the answer would actually be (B). So you can't just stop at the place you stopped; you have to go further.

Instead of doing algebra, I'd translate the given information to:

x < -1

OR

0 < x < 1

For the first situation (x^2 - x) will always be positive.
For the second situation (x^2 - x) will always be negative.

Hope that makes sense! Your thinking on the question might have been totally perfect, by the way, but if you did stop at just that moment, there was still a danger.

-t

[irs031]

When we factor out x^3-x<0, I believe we get x(x^2-1)<0 , not x(x^2-x)<0?
If we have x(x^2-x)<0, then the product would be X^3-X^2<0?

[tommywallach]

You're so right! Oops!

-t