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Math Help - Absolute Values (a non-urgent but roundabout question)

  1. #1
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    Absolute Values (a non-urgent but roundabout question)

    I have a question which is apparently worth 4 marks (of 10!) which states that there is a possible 4 answers for the following:

    Solve |x|(x-1)=2 to which I simply worked out by expanding the LHS to 2x-x=2 and thus x=2

    I have no idea what other way I can look at this, so what am I missing?
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  2. #2
    MHF Contributor red_dog's Avatar
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    If x\geq 0 then x(x-1)=2\Rightarrow x^2-x-2=0\Rightarrow x_1=-1, \ x_2=2.

    But we have to choose the positive root, so x=2

    If x<0 then -x(x-1)=2\Rightarrow -x^2+x-2=0 which has no real roots because the discriminant is negative.
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