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Finding the set of u such that

  1. Sep 12, 2015 #1
    Hi. I'm having trouble understanding the logic behind finding a set of u for u(u-1) > 0. To solve these, I tend to factor as much as possible, then equate each expression to 0 (in this case to > 0) and solve, changing the > to < if I must divide by a negative number. I get u >0 and u > 1 but the solution is given as (-∞,0)U(1,∞), aka u>1 and u<0. Why does it say u<0?
    Any help would be appreciated.

    SOLVED. See below replies + http://tutorial.math.lamar.edu/Classes/Alg/SolvePolyInequalities.aspx
    Last edited: Sep 12, 2015
  2. jcsd
  3. Sep 12, 2015 #2


    Staff: Mentor

    This is not the correct approach. If u(u - 1) > 0 then either
    1) u > 0 AND u - 1 > 0. This leads to u > 0 AND u > 1, which simplifies to u > 1
    2) u < 0 AND u - 1 < 0. This leads to u < 0 AND u < 1, which simplifies to u < 0
    No, (-∞, 0) U (1, ∞) does not mean u > 1 and u < 0, which no number can satisfy. There is no number that is simultaneously greater than 1 and negative.

    (-∞, 0) U (1, ∞) means u < 0 or u > 1. The difference between "and" and "or" here is very significant.
    See above
  4. Sep 12, 2015 #3
    Last edited: Sep 12, 2015
  5. Sep 12, 2015 #4


    Staff: Mentor

    Because if a*b > 0, either both numbers have to be positive or both have to be negative. For example (-2)(-4) = + 8.
  6. Sep 12, 2015 #5
    Alright, so what you have here Is fun.
    in fact when you open the parentheses you will get: u2 - u > 0.

    Now to know when that happends you can simply solve what it should look like.
    since the u^2 has a positive feature you can infer it will be a "smiling" parabule.
    and you can clearly see the u(u-1) You can easly infere that the whole equation will equel zero at u=0 or u=-1

    look at wolf ram alpha and see the discription fits.

    After this you need to look at where the Y axis (the actual u(u-1)) is larger then 0.
    You will find it does at numbers higher then 1 or lower then 0.

    You're approax is deviding it into 2 separate the inequalities and I believe this way is easier and quicker.

    Tell me if I am unclear.
  7. Sep 12, 2015 #6
    @ Spring. I figured it out already thanks to the link posted above. I appreciate the reply though!
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