Results 1 to 6 of 6

Thread: cubic equation help

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    11

    cubic equation help

    show that one fo the roots of the equation $\displaystyle x^3-6x^2+9x-1=0$ is $\displaystyle 4\sin^2\frac{\pi}{18}$ and find the other two roots
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by siegfried View Post
    show that one fo the roots of the equation $\displaystyle x^3-6x^2+9x-1=0$ is $\displaystyle 4\sin^2\frac{\pi}{18}$ and find the other two roots
    Let $\displaystyle x=y+2$.
    This gives, $\displaystyle (y+2)^3 - 6(y+2)^2 + 9(y+2) - 1 = 0$.
    Which simplifies to $\displaystyle y^3-3y+1=0$.

    Let $\displaystyle C = - \frac{\Delta}{108} = - \frac{3}{4}$

    If $\displaystyle A = -\frac{1}{2} + \sqrt{C} = e^{2\pi i/3}$ and $\displaystyle B = -\frac{1}{2} - \sqrt{C} = e^{-2\pi i/3}$.
    Then this means $\displaystyle u=\sqrt[3]{A} = e^{2\pi i/9}$ and $\displaystyle v=\sqrt[3]{B} = e^{-2\pi i/9}$ would produce solutions:
    $\displaystyle u+v,u\zeta + v\zeta^2,u\zeta^2+v\zeta$ (where $\displaystyle \zeta = e^{2\pi i/3})$

    Simplification of these formulas leads to the solutions:
    $\displaystyle 2\cos \frac{2\pi}{9}, 2\cos \frac{8\pi}{9}, 2\cos \frac{14\pi}{9}$.


    And so the three solutions to the original are:
    $\displaystyle x_1 = 2 + 2\cos \frac{2\pi}{9} = 4\sin^2 \frac{\pi}{18}$
    $\displaystyle x_2 = 2 + 2\cos \frac{8\pi}{9} = 4\sin^2 \frac{4\pi}{18}$
    $\displaystyle x_3 = 2 + 2\cos \frac{14\pi}{9}= 4\sin^2 \frac{7\pi}{18}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    11
    thank you theperfecthacker

    i believe you are using cardanos method but i think there is an easier method than cardanos
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by siegfried View Post
    thank you theperfecthacker

    i believe you are using cardanos method but i think there is an easier method than cardanos
    It is Cardano's method with a lot of work omitted as well as computations.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Use the formulas $\displaystyle \cos(2\theta) = 1 - 2\sin^2\theta$ and $\displaystyle \cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. These show that if $\displaystyle x=4\sin^2\theta$ then $\displaystyle \cos(2\theta) = 1 - \tfrac x2$, and so $\displaystyle \cos(6\theta) = 4\left(1 - \tfrac x2\right)^3 - 3\left(1 - \tfrac x2\right) = 1 - \tfrac92x + 3x^2 - \tfrac12x^3$. But if $\displaystyle \theta = \tfrac\pi{18}$ then $\displaystyle \cos(6\theta)=\tfrac12$. That gives the equation $\displaystyle x^3-6x^2+9x-1=0$. The other two solutions will come from the other angles satisfying $\displaystyle \cos(6\theta)=\tfrac12$, namely $\displaystyle \theta = \tfrac{5\pi}{18}$ and $\displaystyle \theta = \tfrac{7\pi}{18}$. (ThePerfectHacker had $\displaystyle \theta = \tfrac{4\pi}{18}$, but I think $\displaystyle \theta = \tfrac{5\pi}{18}$ must be correct.)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Opalg View Post
    (ThePerfectHacker had $\displaystyle \theta = \tfrac{4\pi}{18}$, but I think $\displaystyle \theta = \tfrac{5\pi}{18}$ must be correct.)
    Yes I made a mistake. But we are going to keep it secret.

    The mistake was in the simplification. The solutions that I got to the depressed cubic are correct I just made mistakes in the identities.
    Here is the corrected version:

    1)$\displaystyle 2 + 2\cos \frac{2\pi}{9} = 2 - 2\cos \frac{7\pi}{9} = 4\sin^2 \frac{7\pi}{18}$

    2)$\displaystyle 2 + 2\cos \frac{8\pi}{9} = 2 - 2\cos \frac{\pi}{9} = 4\sin^2 \frac{\pi}{18}$

    3)$\displaystyle 2 + 2\cos \frac{14\pi}{9} = 2 - 2\cos \frac{5\pi}{9} = 4\sin^2 \frac{5\pi}{18}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cubic Equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: Oct 12th 2011, 12:45 AM
  2. Cubic Equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Dec 15th 2009, 04:16 PM
  3. Cubic Equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 21st 2008, 10:45 PM
  4. A cubic equation...
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 12th 2008, 02:05 PM
  5. Cubic Equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Aug 20th 2006, 07:21 AM

Search Tags


/mathhelpforum @mathhelpforum