Results 1 to 6 of 6

Math Help - equations help

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    surrey
    Posts
    4

    equations help

    solve for x+y+z if
    19x^3+95x+57=0
    19y^3+95y+57=0
    19z^3+95z+57=0
    x,yand z are not equal to each other
    Last edited by kahandasa; October 24th 2012 at 07:18 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: equations help

    Quote Originally Posted by kahandasa View Post
    solve for x+y+z if
    19x^3+95x+57=0
    19y^3+95y+57=0
    19z^3+95z+57=0
    All of the equations are identical, so x = y = z. Just solve the first equation for x (if possible) and then you will have y and z as well.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2012
    From
    surrey
    Posts
    4

    Re: equations help

    ya but it is very diificullt. the only way is to play with the eqation and obtain x+y+z
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539

    Re: equations help

    Hello, kahandasa!

    A very strange problem . . .


    \text{Solve for }x+y+z.

    . . \begin{array}{c}19x^3+95x+57\:=\:0 \\ 19y^3+95y+57\:=\:0 \\ 19z^3+95z+57\:=\:0 \end{array}

    x,y\text{ and }z\text{ are not equal to each other.}

    \text{First of all, the equations can be simplified to: }\:\begin{array}{c}x^3 + 5x + 3 \:=\:0 \\ y^3 + 5y + 3 \:=\:0 \\ z^3 + 5z + 3 \:=\:0\end{array}

    \text{Evidently, we have the three distinct roots }x,\,y,\,z
    . . \text{of a cubic equation: }\,u^3 + 0u^2 + 5u + 3 \:=\:0

    \text{Therefore: }\:x+y+z \:=\:0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2012
    From
    surrey
    Posts
    4

    Re: equations help

    i dont understand why x+y+z=0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2012
    From
    Erewhon
    Posts
    159
    Thanks
    104

    Re: equations help

    Quote Originally Posted by kahandasa View Post
    i dont understand why x+y+z=0
    Viete's relations: the sum of the roots of an n-th degree polynomial equation is -1 times the coefficient of the (n-1)-th degree term divided by the coefficient of the n-th degree term. Since there is no term in the square of the variable and we have a cubic the sum of the roots is zero.

    In this case we can be explicit, with soroban's notation:

    (u-x)(u-y)(u-z)=u^3-(x+y+z)u^2+... = u^3+ 0\times u^2+5u+3

    etc..

    Also all the roots are different since by Descartes' rule of signs there are zero positive roots and one negative root, hence exactly one real root, and as this is a polynomial with real coefficients the remaining two complex roots are a conjugate pair and hence not equal.

    CB
    Last edited by zzephod; October 25th 2012 at 05:06 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: April 7th 2010, 02:22 PM
  2. Replies: 3
    Last Post: February 27th 2009, 07:05 PM
  3. Replies: 1
    Last Post: September 1st 2007, 06:35 AM
  4. Replies: 1
    Last Post: July 29th 2007, 02:37 PM
  5. Replies: 3
    Last Post: July 9th 2007, 05:30 PM

Search Tags


/mathhelpforum @mathhelpforum