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Math Help - An equation with 6 roots

  1. #1
    Member Jones's Avatar
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    An equation with 6 roots

    Hi, how do i do when i want to solve an equation with 6 roots.

    I have prevously solved equations with 4 roots Z^n = -i.

    But this time im stuck. The equation is looking like this: Z^6=-64
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  2. #2
    Forum Admin topsquark's Avatar
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    A couple of ways. This is how I would do it:

    z^6 + 64 = 0
    This is the sum of two perfect cubes, (z^2)^3 and 4^3 so this factors:

    (z^2 + 4)(z^4 - 4z^2 + 16) = 0

    So set both factors equal to 0:
    z^2 + 4 = 0 gives z = (+/-) 2i

    The second factor gives:
    z^4 - 4z^2 + 16 = 0, which can be "solved" via the quadratic formula:

    z^2 = 2 (+/-) 2i*sqrt(3)

    The best way to continue from here is to convert to complex polar coordinate form:
    2 + 2i*sqrt(3) = 4 exp(i*pi/3)
    2 - 2i*sqrt(3) = 4 exp(-i*pi/3)

    So taking the square root of both of these gives:
    z = (+/-) 2 exp(i*pi/6) and z = (+/-) 2 exp(-i*pi/6)

    Or if you want these in "rectangular" form:
    z = (+/-) (sqrt(3) + i) and z = (+/-)(sqrt(3) - i)

    (And don't forget the z = (+/-) 2i)

    -Dan
    Last edited by topsquark; September 27th 2006 at 09:37 AM.
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  3. #3
    Member Jones's Avatar
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    I didn't know it was possible to solve equations that hade more than four roots
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jones View Post
    I didn't know it was possible to solve equations that hade more than four roots
    Yeah, well, since my factoring abilities today resemble those of a drunken donkey (insert appropriate cuss word here) I almost couldn't do it either!

    There is a theorem that says you cannot solve a general nth degree polynomial equation for n > 4 (using a formula), but specific polynomials might be solvable using specific techniques. This is one of them.

    -Dan
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Jones View Post
    Hi, how do i do when i want to solve an equation with 6 roots.

    I have prevously solved equations with 4 roots Z^n = -i.

    But this time im stuck. The equation is looking like this: Z^6=-64
    -1=exp(i pi (1+2k)), k=0, 1, 2, ..

    so

    z^6=64 exp(i pi (1+2k)), k=0, 1, 2, ..

    so:

    z=(64)^(1/6) exp(i pi/6(1+2k)) k=0, 1, 2..

    but there are only 6 distinct values of this corresponding to the
    6-th roots of -1, exp(i pi/6(1+2k)) k=0, 1, 2, 3, 4, 5.

    RonL
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  6. #6
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    Quote Originally Posted by Jones View Post
    I didn't know it was possible to solve equations that hade more than four roots
    Actually, your equation is not much of an equation; I would consider it more a matter of solving for the sixth root of a variable, rather than of an equation.

    When I think about finding the roots of an equation, I tend to think of an equation that looks as follows:

    a*x^4 + b*x^3 + c*x^2 + d*x + e = 0

    For an equation like this, a fourth-degree equation, there are formulae for calculating exact solutions. Like Dan said, finding exact solutions for higher degree equations is not, in general, possible. However, numerical techniques exist. In fact, there are several numerical algorithms, such as Jenkins-Traub, that give excellent results for equations up to degree 100 or higher. Eigenvalue techniques can also be used.
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