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Math Help - Real and non-real roots to a quintic equation

  1. #1
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    Real and non-real roots to a quintic equation

    The question is:

    Establish that alpha=2+3i is a root of the quintic equation

    3z^5 + 2z^4 - 64z^3 +384z^2 - 667z +182 =0

    and find the remaining roots.

    Now, I know that as it's a quintic there will be five roots. As I'm told that 2+3i is a root, I can know that 2-3i is also a root. And once I get so far I will be able to divide, take the one real root and the other two complex roots.

    But I'm falling at the first hurdle, how do I show that 2+3i is a root?
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  2. #2
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    Quote Originally Posted by MickHarford View Post
    The question is:

    Establish that alpha=2+3i is a root of the quintic equation

    3z^5 + 2z^4 - 64z^3 +384z^2 - 667z +182 =0

    and find the remaining roots.

    Now, I know that as it's a quintic there will be five roots. As I'm told that 2+3i is a root, I can know that 2-3i is also a root. And once I get so far I will be able to divide, take the one real root and the other two complex roots.

    But I'm falling at the first hurdle, how do I show that 2+3i is a root?
    If 2 + 3i and 2 - 3i are two solutions, what quadratic do they represent?

    Once you obtain that and factor it out, the cubic will factor nicely.

    To show 2 + 3i is a root, plug it in.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    If 2 + 3i and 2 - 3i are two solutions, what quadratic do they represent?

    Once you obtain that and factor it out, the cubic will factor nicely.

    To show 2 + 3i is a root, plug it in.
    This is the part I don't get, or at least I'm being thick.



    I've got an example written down:
    p(z) = z^4 - 4z^3 +55z^2 -8z +106
    2+7i and 2-7i are roots.
    So p(z)=(z^2-4z+53)(z^2+2)

    I understand how the division and eventual solutions are obtained, I jsut don't quite understand the step to what the quadratic from the solutions is?

    It's probably something incredibly simple isn't it?
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  4. #4
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    Quote Originally Posted by MickHarford View Post
    This is the part I don't get, or at least I'm being thick.



    I've got an example written down:
    p(z) = z^4 - 4z^3 +55z^2 -8z +106
    2+7i and 2-7i are roots.
    So p(z)=(z^2-4z+53)(z^2+2)

    I understand how the division and eventual solutions are obtained, I jsut don't quite understand the step to what the quadratic from the solutions is?

    It's probably something incredibly simple isn't it?
    What would z need to be is the question we are asking:

    \displaystyle z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\cdots

    \displaystyle 2\pm 3i=\frac{-(-4)\pm\sqrt{-36}}{2}=\frac{-(-4)\pm\sqrt{16-4ac}}{2(1)}

    What are a and c?
    Last edited by dwsmith; February 13th 2011 at 10:22 AM. Reason: Changed x to z
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  5. #5
    Senior Member roninpro's Avatar
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    Try multiplying (z-[2-3i])(z-[2+3i]).
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    Quote Originally Posted by roninpro View Post
    Try multiplying (z-[2-3i])(z-[2+3i]).
    I got z^2 - 4z + 13 can someone verify please?

    I'll crack on with the question now anyway!
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  7. #7
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    Quote Originally Posted by MickHarford View Post
    I got z^2 - 4z + 13 can someone verify please?

    I'll crack on with the question now anyway!
    that's fine.
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  8. #8
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    Thanks guys. Solved the problem, after my initial struggles it just required two divisions through.

    Solution set I got was 2+3i, 2-3i, 2, 7 and -1/3 (going off memory right now)
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  9. #9
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    Quote Originally Posted by roninpro View Post
    Try multiplying (z-[2-3i])(z-[2+3i]).
    Easier to expand ([z - 2] + 3i)([z - 2] - 3i).
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