Results 1 to 4 of 4

Math Help - Urgent: 4 degree polynomial with irrational roots

  1. #1
    Newbie
    Joined
    Nov 2006
    Posts
    9

    Urgent: 4 degree polynomial with irrational roots

    Hi

    which specific formula/method I can use to find the complex roots of a polynomial of degree 4 on form:

    p(x) = x^4 + a x^3 + bx^2 + ax + a_0 = 0


    which has irrational coefficients
    a = -1 - \sqrt(2) and b =  2+ \sqrt(2) and

    where a_0 = 1


    If I write the equation as x^4-x^3+2x^2-x+1-\sqrt2(x^3-x^2+x)=0, then you will notice that it factorises as (x^2-x+1)(x^2-\sqrt2x+1)=0

    Then by fundamental theorem of algebra I see that

    p(x) = (x-(\frac{1}{2}) + \frac{\sqrt(3)}{2}i)) (x-(\frac{1}{2}) - \frac{\sqrt(3)}{2}i)) \cdot ((x - (\frac{\sqrt(2)}{2} + \frac{\sqrt(2)}{2} i)) ((x - (\frac{\sqrt(2)}{2} - \frac{\sqrt(2)}{2} i)) = (x^2-x+1)(x^2-\sqrt2x+1) = x^4-x^3+2x^2-x+1-\sqrt2(x^3-x^2+x)

    where the complex roots of the original polymial p(x) is


    x = ((\frac{1}{2}) \pm \frac{\sqrt(3)}{2}i), (\frac{\sqrt(2)}{2} \pm \frac{\sqrt(2)}{2} i)


    Getting back to the original polynomial
    (*) p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0

    where a,b \in \mathbb{C}

    I would like to prove that a complex number x makes (*) true iff

    s = x + x^{-1} is a root of the Q(s) = s^2 + as + (b-2)

    I see that that Q(x + x^{-1}) = \frac{p(x)}{x^2}

    So the solutions of in my equation s^2+as+(b-2)=0 are s=\frac{-a\pm\sqrt{a^2-4(b-2)}}{2}.

    I then need to plug each of these numbers into the equation s=x+x^{-1}, which can be written as x^2-sx+1=0. Using the quadratic formula again, you get x=\frac{s\pm\sqrt{s^2-4}}{2}.

    By the way do x^-1 then exist ?
    Yes, there's no problem about that. In fact since s=x+x^{-1}, you can see that x^{-1}=s-x


    I get this x value for both values of s to be:

     x =\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4}

    This expression makes the equation

    p(x) = \frac{p(x)}{x^2} true

    since p(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4}) = \newline \frac{p(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4})}{(\frac{\sqrt{-2(a \sqrt{a^2 - 4 \cdot (b-2) - a^2 + 2(b+2)}} + \sqrt{a^2 - 4 (b-2) -a}}{4})^2} = 0

    This proves that there exist a number x which is both a root of the original polynomial and Q(s = x + x^{-1}), since p(x) = \frac{p(x)}{x^2}.

    But one question remains on my part. Since x is supposedly a complex number. Is it a complex number in its current form? That I'm a bit unsure of. Since then it should be written in the form x = a + bi ?

    Best Regards,
    Billy
    Last edited by Billy2007; December 10th 2006 at 02:02 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,923
    Thanks
    332
    Awards
    1
    If you want to call x a complex number, it's a complex number. There's no need to write it in a + bi form, though you may do so if you like.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    You could try running the particular form of quartic that you have through
    the quartic formula and see what you get.

    Background on the formula can be found here.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I know how to solve the cubic algebraically but never learned how to solve the quartic, it is too long. I find it amazing how these Italian mathematicians were so good at manipulation of radical terms and factorizations.

    I read that one Italian mathematician was killed for being able to solve the quartic. Because people believed it was not withing human skills and thus he made a pact with a Devil (similar story with the violinist, Pagannini, they used the same excuse).

    This is my 37th Post!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] roots of a 5-degree polynomial
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 6th 2011, 08:47 AM
  2. [SOLVED] Roots of a Sixth-Degree Polynomial
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 27th 2011, 02:53 AM
  3. roots of a 3rd degree polynomial.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 10th 2010, 12:29 PM
  4. Replies: 3
    Last Post: April 7th 2010, 02:51 AM
  5. Replies: 1
    Last Post: February 8th 2010, 03:50 AM

Search Tags


/mathhelpforum @mathhelpforum