Expert: Steve Holleran - 7/28/2008

Question
I'm trying to solve Problem # 29 of this Website: http://m.njit.edu/Undergraduate/placement.pdf I Thank you In advance just for your desire to help others. Thank you.


Answer
Hi Jocsan,

This one is not easy, but try the following:

There is a common factor of (x^2 - x)^(-1/3) on the left, so let's factor it out:

  (x^2 - x)^(-1/3) * [(2x(2x-1))/3 + (x^2 - x)] = 0


  (x^2 - x)^(-1/3) * [(4x^2 - 2x)/3 + (x^2 - x)] = 0

  (x^2 - x)^(-1/3) * [4/3 * x^2 - 2/3 * x + 1x^2 - 1x] = 0

Combine terms the bracket:

  (x^2 - x)^(-1/3) * [ 7/3 * x^2 - 5/3 * x] = 0

Factor 1/3 * x out on the inside of the bracket and re-write:

     [1/3 * x (7x - 5)] / (x^2 - x)^(1/3) = 0

Setting the top = 0 gives x = 0, which is ruled out at the outset,
and x = 5/7.   I did a check on this, and it was not easy, but I believe it does check out.

Good luck!
Steve