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Explanation for Simultaneous Equations Solution

  1. Mar 10, 2012 #1
    Hi, there's this simultaneous equation problem, however I already know the solution, I just want to know more about the solution.
    a+b+c=2,bc+ac+ab=-5, abc=-6
    The solution is:
    Consider the polynomial (x-a)(x-b)(x-c), where a,b,c are constants.
    (x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(bc+ac+ab)x-abc
    Substitute the given equations into the polynomial:
    (x-a)(x-b)(x-c) = x^3-2x^2-5x+6
    (x-a)(x-b)(x-c) = (x-1)(x+2)(x-3)
    Therefore, a,b,c = 1,-2,3 in any order.

    My question is, how can you know that the three equations relate to a cubic purely from inspection? And is this some broader strategy to solve non-linear simultaneous equations?

    If the normal method of substituting is used to solve these simultaneous equations,
    a=2-b-c
    sub a in both equations:
    b(2-b-c)+c(2-b-c)+bc=-5
    b^2+bc+c^2-2b-2c=5
    and
    (2-b-c)bc=-6
    bc(b+c-2)=6
    b^2c+bc^2-2bc=6
    There isn't much else you can do, unless get b+c-2 as the subject of both equations, then sub again, which doesn't yield anything, though it's not surprising that you can't solve this by normal methods, as if you could it would mean you could factorise any polynomial of degree 3, and by extension, of any degree easily, which you can't.

    Also, just investigating, if you do a similar thing for a quadratic:
    a+b=-3
    ab=2
    Again we can see see these are the coefficients of a polynomial, namely a quadratic:
    (x-a)(x-b) = x^2-(a+b)x+ab
    sub the given equations in:
    =x^2+3x+2
    =(x+1)(x+2)
    so a, b=-1,-2 in any order
    However, we can also solve this through normal methods:
    a=2/b
    2/b+b=-3
    2+b^2=-3b
    b^2+3b+2=0
    (b+1)(b+2)=0
    b=-1,-2, then solve for a
    This could suggest you could do something similar for the cubic, however I can't see how.
     
  2. jcsd
  3. Mar 10, 2012 #2

    epenguin

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    Homework Helper
    Gold Member

    For the first question, they are just functions ('symmetric functions of roots') which become very familiar as soon as you study theory of algebraic equations. In a fairly obvious notation related to yours they are Ʃa, Ʃab, Ʃabc and for higher than cubic it goes on Ʃabcd etc.

    For the second question the answer is no. You can do it in this case because there are nice whole numbers that guide you to a solution. e.g. abc = -6 = -1X3X2 = 1X-2X3 = ... = -1X1X6 =... - the whole number possibilities are limited. There is a nice answer to your natural question (it looks like you ought to be able to do it and then it always eludes you and seems like pinning mercury) on p 36-37 of Burnside and Panton "The Theory of Equations" vol 1 a bit too long for me to reproduce right now, and no doubt other algebra books. Crudely I might say that because they are symmetric functions there is nothing in them that distinguishes one root from another, there is nothing that distinguishes your a and your b etc., therefore you cannot get one and then from that get another, you would have to be able to get them all by the same procedure and that, except in special vases and although it is not self-evident, you cannot do.
     
    Last edited: Mar 10, 2012
  4. Mar 10, 2012 #3
    Thanks for your answer.
     
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