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Math Help - Find the values of x

  1. #1
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    Find the values of x

    such that  -2 < x^3 - 2x^2 + x - 2 < 0 .
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  2. #2
    MHF Contributor alexmahone's Avatar
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    -2<x^3-2x^2+x-2<0

    -2<(x-2)(x^2+1)<0

    x^2+1 is positive

    Dividing throughout by x^2+1,

    -2<(x-2)<0

    0<x<2
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by mark1950 View Post
    such that  -2 < x^3 - 2x^2 + x - 2 < 0 .
    Concentrate on one side of the inequality at a time.

    -2 < x^3-2x^2+x-2

    \implies\ 0<x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2

    Solive this, then go on to the next inequality.

    x^3-2x^2+x-2<0

    Solve this (hint: x-2 is a factor of the LHS). Then combine the two solutions.

    Quote Originally Posted by alexmahone View Post
    -2<(x-2)(x^2+1)<0

    x^2+1 is positive

    Dividing throughout by x^2+1,

    -2<(x-2)<0
    -2 divided by x^2+1 is not -2.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    -deleted-
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by TheAbstractionist View Post


    -2 divided by x^2+1 is not -2.
    I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.
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  6. #6
    Senior Member TheAbstractionist's Avatar
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    I mean if you divide

    -2<(x-2)(x^2+1)<0

    by x^2+1 you should get

    \frac{-2}{x^2+1}<x-2<0

    not -2<x-2<0.

    In any case, x=1 is not a solution.
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  7. #7
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    Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?
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  8. #8
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mark1950 View Post
    Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?
    That's because x=1 is NOT a solution.
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  9. #9
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    Quote Originally Posted by alexmahone View Post
    I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.
    Think about this example and you'll see you're in error:
    0<2x<1
    We can all agree that 2 is positive so by your working:
    0<x<1
    Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term.
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  10. #10
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by pfarnall View Post
    Think about this example and you'll see you're in error:
    0<2x<1
    We can all agree that 2 is positive so by your working:
    0<x<1
    Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term.
    Fair enough. Sorry.
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