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Math Help - inequality

  1. #1
    Senior Member
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    inequality

    Find the solutions set of the inequality |x-2|<\frac{1}{x} , where x is not 0 .



    THanks .
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    it's quite obvious that x\ne0, that's an assumption that it must be implicit and that you should know.

    well then, by a well known property your inequality equals -\frac{1}{x}<x-2<\frac{1}{x}, so you have to solve each one.

    i'll help ya with the first one:

    \begin{aligned}<br />
   x-2&>-\frac{1}{x} \\ <br />
  x-2+\frac{1}{x}&>0 \\ <br />
  \frac{x^{2}-2x+1}{x}&>0, <br />
\end{aligned}

    the numerator is a perfect square, but it has a critical point at x=1, but we don't actually want that, since that'd turn the numerator zero and the inequality won't be satisfied, so to solve the previous inequality we just require that x>0, but since we've stated that x\ne1, the first solution set is (0,1)\cup(1,\infty).

    apply the same procedure for the second inequality, that'd yield a second solution set, which you need to intersect with the first one to get the full solution set.
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  3. #3
    MHF Contributor red_dog's Avatar
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    If x<0, the inequality has no solution.

    If x>0 the inequality is x|x-2|<1

    1) x\in(0,2)\Rightarrow x(2-x)<1\Rightarrow (x-1)^2>0, which is true.

    2) x\in[2,\infty)\Rightarrow x^2-2x-1<0\Rightarrow x\in(1-\sqrt{2},1+\sqrt{2})\cap [2,\infty)\Rightarrow x\in[2,1+\sqrt{2})

    From 1) and 2), x\in(0,2)\cup[2,1+\sqrt{2})\Rightarrow x\in(0,1+\sqrt{2})
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