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Thread: Roots of nth order polynomials

  1. #1
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    Roots of nth order polynomials

    I'm having a heck of a time with a few sections on my differential equations homework. Solving nth order polynomials is only part of each problem, and I'm having enough trouble with that. There seems to be no method for solving these other than guess and check, and when most of them end up having complex roots it feels like it's impossible.

    For example:

     r^4 + 2r^2 +1 = 0

    How are you supposed to solve for the roots of this? As far as I can tell they're complex, and I don't think I can guess and check =\ Thanks for any help guys
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  2. #2
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    r^4  + 2r^2  + 1 = \left( {r^2  + 1} \right)^2  = 0\quad  \Rightarrow \quad r = i\,\& \, - i
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  3. #3
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    Just factoring we get:

    (r^{2}+1)^{2}

    The roots are clearly i and -i
    Last edited by galactus; Jul 2nd 2008 at 03:58 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by griffsterb View Post
    I'm having a heck of a time with a few sections on my differential equations homework. Solving nth order polynomials is only part of each problem, and I'm having enough trouble with that. There seems to be no method for solving these other than guess and check, and when most of them end up having complex roots it feels like it's impossible.

    For example:

     r^4 + 2r^2 +1 = 0

    How are you supposed to solve for the roots of this? As far as I can tell they're complex, and I don't think I can guess and check =\ Thanks for any help guys
    Alternatively

    Let r^2=\psi

    So we have

    \psi^2+2\psi+1=0

    By the quadractic formula

    \psi=\frac{-2\pm\sqrt{4-4(1)(1)}}{2}\Rightarrow{\psi=-1}

    So

    \psi=-1\Rightarrow{r^2=-1}\Rightarrow{r=\pm{i}}
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