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Math Help - Complex Roots of Polynomial

  1. #1
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    Complex Roots of Polynomial

    P(x) = x^4 + cx^2 + e, where c and e are real

    If sqrt3 + i is a root, determine c and e, then factor into quadratic factors with linear factors

    Here's my attempt:
    Since c and e are real,
    then sqrt3 - i is a root (complex conjugates theorem)

    Also
    P(-x) = P(x)
    therefore P(x) is even

    Hence
    -(sqrt3 + i) and -(sqrt3 - i) are also roots

    from there i got e = 16 (product of roots)
    c= ...

    However the answers have it differently.
    It subbed sqrt3 + i and sqrt3 - i into P(x), then solved simultaneously acquiring:
    e= - 8, c = 8

    Can someone explain the flaw in my reasoning? Or perhaps answers is wrong?
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  2. #2
    A Plied Mathematician
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    I agree with your reasoning. I double-checked the book's coefficients in Mathematica. At least one of the roots Mathematica found for the book's coefficients is real!?! But I totally agree that you can find all the roots using conjugates and the function's evenness. So, I would ignore the book's answer.

    Cheers!
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  3. #3
    MHF Contributor chisigma's Avatar
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    P(*) is a real coefficients polynomial that contains only the even powers of x, so that if \sigma + i\ \omega is a root of P(*), \sigma - i\ \omega , - \sigma + i\ \omega and -\sigma - i\ \omega are also roots od P(*) and is...

     P(x) = (x -  \sigma - i\ \omega)\ (x -  \sigma + i\ \omega)\ (x +  \sigma - i\ \omega)\ (x +  \sigma + i\ \omega) (1)

    Now what You have to do is setting in (1) \sigma =  \sqrt{3} and \omega = 1 and imposing that...

    P(x) =  x^{4} + c\ x^{2} + e (2)

    Kind regards

    \chi \sigma
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