Results 1 to 4 of 4

Math Help - Roots of cubic equation

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    145

    Roots of cubic equation

    Hi,

    Kindly help me with this roots question...please

    If the roots of the equation 4x + 7x - 5x - 1 = 0 are α, β and γ, find the equation whose roots are
    i. α + 1, β + 1 and γ + 1

    ii. α, β, γ
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    We have x_1=\alpha, \ x_2=\beta, \ x_3=\gamma. Then

    x_1+x_2+x_3=-\frac{7}{4}

    x_1x_2+x_1x_3+x_2x_3=-\frac{5}{4}

    x_1x_2x_3=\frac{1}{4}

    1) y_1=x_1+1, \ y_2=x_2+1, \ y_3=x_3+1

    In this case the easiest way is to let y=x+1. Then x=y-1 and replace x in the equation:

    4(y-1)^3+7(y-1)^2-5(y-1)-1=0

    Now continue.

    2) Let S_1=y_1+y_2+y_3=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)

    S_2=y_1y_2+y_1y_3+y_2y_3=x_1^2x_2^2+x_1^2x_3^2+x_2  ^2x_3^2=

    =(x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)

    S_3=y_1y_2y_3=(x_1x_2x_3)^2

    Then the equation is y^3-S_1y^2+S_2y-S_3=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Quote Originally Posted by red_dog View Post
    We have x_1=\alpha, \ x_2=\beta, \ x_3=\gamma. Then

    x_1+x_2+x_3=-\frac{7}{4}

    x_1x_2+x_1x_3+x_2x_3=-\frac{5}{4}

    x_1x_2x_3=\frac{1}{4}

    1) y_1=x_1+1, \ y_2=x_2+1, \ y_3=x_3+1

    In this case the easiest way is to let y=x+1. Then x=y-1 and replace x in the equation:

    4(y-1)^3+7(y-1)^2-5(y-1)-1=0

    Now continue.

    2) Let S_1=y_1+y_2+y_3=x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)

    S_2=y_1y_2+y_1y_3+y_2y_3=x_1^2x_2^2+x_1^2x_3^2+x_2  ^2x_3^2=

    =(x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)

    S_3=y_1y_2y_3=(x_1x_2x_3)^2

    Then the equation is y^3-S_1y^2+S_2y-S_3=0

    I'm having trouble with similar problems. How do you get:

    x_1+x_2+x_3=-\frac{7}{4}

    ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Quote Originally Posted by adkinsjr View Post
    I'm having trouble with similar problems. How do you get:

    x_1+x_2+x_3=-\frac{7}{4}

    ??
    If x_1, \ x_2, \ x_3 are the roots of equation ax^3+bx^2+cx+d=0 then

    x_1+x_2+x_3=-\frac{b}{a}

    x_1x_2+x_1x_3+x_2x_3=\frac{c}{a}

    x_1x_2x_3=-\frac{d}{a}

    These are the relations of Viete.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Forming a Cubic Equation from Roots
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: September 13th 2010, 04:31 PM
  2. Roots of cubic equations and new equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 13th 2010, 03:46 AM
  3. Roots of cubic equations and new equation
    Posted in the Algebra Forum
    Replies: 11
    Last Post: January 4th 2010, 02:49 PM
  4. Roots of a cubic equation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 3rd 2010, 01:59 PM
  5. Replies: 4
    Last Post: October 16th 2008, 04:46 AM

Search Tags


/mathhelpforum @mathhelpforum