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Math Help - [SOLVED] Polynomial x and i

  1. #1
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    [SOLVED] Polynomial x and i

    Hi
    I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25=0
    but I don't get the answer. I used quadratic X(3x^2-11x+25)-25=0 but that didn’t work.

    Do I need to use rational roots test and long division to solve this?
    Last edited by Neverquit; April 14th 2010 at 02:38 AM. Reason: =0
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  2. #2
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    Quote Originally Posted by Neverquit View Post
    Hi
    I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25
    but I don't get the answer. I used quadratic X(3x^2-11x+25)-25=0 but that didnít work.

    Do I need to use rational roots test and long division to solve this?

    Or you'll have to use the method by Tartaglia- del Ferro, which is a very nasty thing, but sometimes there's no other way: Cubic function - Wikipedia, the free encyclopedia

    Tonio
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  3. #3
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    Your question, as written, "I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25" doesn't make any sense. x can be any complex number in that formula. Tonio assumed that you meant to solve 3x^3- 11x^2+ 25x-25= 0. Is that correct?
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  4. #4
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    correction

    Equation is suppose to be: 3x^3-11x^2+25x-25=0
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  5. #5
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    Cardanos method for polynomials

    Cardanos method (below) looks like the easiest way.
    I just have to divide 3x^3-11x^2+25x-25 by the leading coefficient 3 to get x^3-(11/3)x^2+(25/3)x-(25/3)=0.
    That gives a=(-11/3),b=(25/3),c=(-25/3).Then solve for p,q,u.

    Should this work??

    For
    the solutions for x are given by

    where


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  6. #6
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    Cardanos method for polynomials

    Cardanos method gave me the following complex root solutions:
    x=2.8444i+1.222

    and

    x=-1.7175i+1.222

    These solutions look close to the answer 1+2i and 1-2i.... just need to find the real root of 5/3.
    To find the real root of 5/3 should I use Cardanos method or real roots test?
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  7. #7
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    Perhaps we shouldn't be so quick to jump to Tartaglia.

    Actually, it would have been far simpler to do what you originally suggested: "use rational roots test and long division".

    The rational roots test shows that only 5, -5, 5/3, and -5/3 could be rational roots. Trying those, we find that 5/3 is, in fact, a root. That means that (3x- 5) is a factor. Dividing 3x^3- 11x^2+ 25x- 25 by 3x- 5 gives a quotient of x^2- 2x+ 5 with no remainder.

    3x^3- 11x^2+ 25x- 25= (3x- 5)(x^2- 2x+ 5)

    x^2- 2x+ 5= x^2- 2x+ 1+ 4= (x- 1)^2+ 4= 0 so
    (x- 1)^2= -4 and x- 1= \pm 2i, x= 1\pm 2i.

    The roots of the original equation are 5/3, 1+ 2i, and 1- 2i.

    (Though, I am, frankly, impressed that you got such accurate roots using Cardano's method.)
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