Results 1 to 7 of 7

Math Help - Forming a Cubic Equation from Roots

  1. #1
    Member alexgeek's Avatar
    Joined
    Mar 2010
    From
    Wales, UK
    Posts
    77

    Forming a Cubic Equation from Roots

    Stuck on a further pure question, asked one of my maths teachers today but she hadn't done most of the paper in 20 years.
    Here's the question:
    Given that the roots of the cubic equation
    x^3+4x^2+3x+2=0
    are \alpha, \beta, \gamma determine the cubic equation with roots \beta \gamma , \gamma \alpha , \alpha \beta

    All I have so far is
    \alpha + \beta + \gamma = -4 \: ; \: \beta \gamma + \gamma \alpha + \alpha \beta = 3 \: ; \: \alpha \beta \gamma = -2

    I assume the new equations will looks like:
    x^3-(\beta \gamma , \gamma \alpha , \alpha \beta)x^2+(\alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma ^2)x - \alpha^2 \beta^2 \gamma^2

    I'm pretty sure the coefficient of the x^2 term will be \beta \gamma + \gamma \alpha + \alpha \beta which equals 3.

    And for the constants:
    \alpha \beta \gamma = -2
    (\alpha \beta \gamma)^2 = 4
    \alpha^2 \beta^2 \gamma^2 - 2(\beta \gamma + \gamma \alpha + \alpha \beta)=4
    \alpha^2 \beta^2 \gamma^2 -2 \times 3 = 4
    \alpha^2 \beta^2 \gamma^2 = 10

    I think I'm correct so far (may be some typos though), but I don't know how to get the x term.

    Thanks.


    edit

    I got it, just factoring the coefficients of the x term:
    \alpha \beta \gamma (\alpha + \beta + \gamma)
    brain ache now.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    I get

    x^3+4x^2+3x+2 = (x-\alpha)(x-\beta)(x-\gamma)

    x^3+4x^2+3x+2 = x^3-(\beta+\alpha+\gamma)x^2+(\alpha\beta-\gamma\beta-\alpha\gamma)x-\gamma\alpha\beta

    so

    4 = -(\beta+\alpha+\gamma), 3=\alpha\beta-\gamma\beta-\alpha\gamma, 2 =-\gamma\alpha\beta
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member alexgeek's Avatar
    Joined
    Mar 2010
    From
    Wales, UK
    Posts
    77
    Quote Originally Posted by pickslides View Post
    I get

    x^3+4x^2+3x+2 = (x-\alpha)(x-\beta)(x-\gamma)

    x^3+4x^2+3x+2 = x^3-(\beta+\alpha+\gamma)x^2+(\alpha\beta-\gamma\beta-\alpha\gamma)x-\gamma\alpha\beta

    so

    4 = -(\beta+\alpha+\gamma), 3=\alpha\beta-\gamma\beta-\alpha\gamma, 2 =-\gamma\alpha\beta
    Yeah that's what I got, the new equation I formed was:
    <br />
x^3-3x^2 +8x + 1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, alexgeek!

    You're wandering a bit, but I'd say you got it (sort of).


    \text{Given that the roots of the equation }x^3+4x^2+3x+2\:=\:0\,\text{ are }\,a, b, c
    \text{determine the cubic equation with roots: }\:ab, bc, ac

    \text{All I have so far is: }\;\begin{array}{ccc} a + b + c &=& \text{-}4 \\<br />
ab + bc + ac &=& 3 \\<br />
abc &=&\text{-}2 \end{array}


    \text{I assume the new equation will looks like:}
    . . x^3-(ab + bc + ac)x^2 + (a^2bc+ ab^2c + abc^2)x - a^2b^2c^2 \;=\;0

    \text{The coefficient of the }x^2\text{ term will be }(ab + bc + ac) \:=\:3

    \text{And the constant: }\:abc \:=\:-2 \quad\Rightarrow\quad a^2b^2c^2 \:=\:4

    The coefficient of the \,x-term is:

    . . a^2bc + ab^2c + abc^2 \:=\:abc(a+b+c) \;=\; (-2)(-4) \:=\:8


    Therefore, the cubic equation is: . x^3 - 3x^2 + 8x - 4 \:=\:0

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    What were your solutions for  \gamma ,\alpha ,\beta ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member alexgeek's Avatar
    Joined
    Mar 2010
    From
    Wales, UK
    Posts
    77
    The individual roots? The question doesn't ask for them, they're some decimal complex numbers and one decimal real solution, http://www.wolframalpha.com/input/?i=x^3%2B4x^2%2B3x%2B2%3D0
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member alexgeek's Avatar
    Joined
    Mar 2010
    From
    Wales, UK
    Posts
    77
    Actually I think the equation is <br />
x^{3}-3x^{2}+(-2)(-4)x-(2)^{2}\equiv x^{3}-3x^{2}+8x-4<br />

    Messed up the constants I think.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Quadratic and cubic equation -show that -(common roots)
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 19th 2011, 11:47 PM
  2. Roots of cubic equations and new equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 13th 2010, 02:46 AM
  3. Roots of cubic equations and new equation
    Posted in the Algebra Forum
    Replies: 11
    Last Post: January 4th 2010, 01:49 PM
  4. Roots of a cubic equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 3rd 2010, 12:59 PM
  5. Roots of cubic equation
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 11th 2009, 05:35 AM

Search Tags


/mathhelpforum @mathhelpforum