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Math Help - Values of a

  1. #1
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    Values of a

    For what values of a there are exactly two real numbers that satify the equation:

    x^3 + (a-1)x^3 + (9-a)x - 9 = 0 ?
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  2. #2
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    x = { 1, \frac{1}{2}\,{\frac {-a+\sqrt {{a}^{2}-36\,a}}{a}}, \frac{1}{2}\,{\frac {-a-\sqrt {{a}^{2}-36\,a}}{a}}}

    So, 1 is a real number. Now pick an 'a' such that the 2nd and 3rd element are equal.

    a = 36

    Note: I only checked the 'a' in the real. There might be other ways of satisfying the question.
    Last edited by paultwang; May 23rd 2005 at 08:37 AM.
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  3. #3
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    oops

    Sorry but there's a typo, It should read

    x^3 + (a-1)x^2 + (9-a)x - 9 = 0

    Instead of:

    x^3 + (a-1)x^3 + (9-a)x - 9 = 0

    Thanks!
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  4. #4
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     x = \{ 1 , -\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} , -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36} \}

    Make the 2nd and 3rd element equal:

    -\frac{1}{2}\,a+\frac{1}{2}\,\sqrt {{a}^{2}-36} = -\frac{1}{2}\,a-\frac{1}{2}\,\sqrt {{a}^{2}-36}

     a = \{6, -6\}
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  5. #5
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    Thanks a lot paultwang!
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