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Thread: 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 $\displaystyle |x|(x-1)=2$ to which I simply worked out by expanding the LHS to $\displaystyle 2x-x=2$ and thus $\displaystyle x=2$

    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 $\displaystyle x\geq 0$ then $\displaystyle x(x-1)=2\Rightarrow x^2-x-2=0\Rightarrow x_1=-1, \ x_2=2$.

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

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