1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving a cubic equation

  1. Aug 22, 2010 #1
    1. The problem statement, all variables and given/known data

    I've been given the following quartic equation in a question:

    [tex]x = (1 + 2a + a^2)x - (2a + 3a^2 + a^3)x^2 + (2a^2 + 2a^3)x^3 - a^3x^4[/tex]

    and need to show that two of the solutions (that aren't 0 and 1) can be given by:

    [tex] x= \frac{(2+a) \pm \sqrt{a^2 - 4}}{2a}[/tex]

    2. The attempt at a solution

    I think I have to start by dividing it all by x - when I put this new equation into wolfram alpha it gives the right answer, I just have no idea how to get there (or where to start), since I haven't done a whole lot with cubics or quartics before.
     
    Last edited: Aug 22, 2010
  2. jcsd
  3. Aug 22, 2010 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Can you say why you think so?
     
  4. Aug 22, 2010 #3

    Borek

    User Avatar

    Staff: Mentor

    Why don't you just put expression given as x into the equation?
     
  5. Aug 22, 2010 #4
    Sorry, I left out a part of the equation above. It was actually x = [what I already gave], I've edited that in now. It just seems like that way it would be easier to rearrange into the form needed, I'm not 100% sure about that though.
     
  6. Aug 22, 2010 #5

    Borek

    User Avatar

    Staff: Mentor

    I thought you mean f(x) = 0. Still, what I wrote above holds.
     
  7. Aug 22, 2010 #6
    what did you mean by putting the expression given as x into the equation? Putting it into the second one there/the other way around?
     
  8. Aug 22, 2010 #7

    Borek

    User Avatar

    Staff: Mentor

    Sorry if my English failed.

    Substitute value of x given in the second equation for all occurrences of x in the first equation. That will yield equation in "a" - and if second equation really gives solution, equation in "a" should be just an identity.
     
  9. Aug 22, 2010 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are given two putative solutions. Just put them into the equation, do the algebra, and see whether or not they satisfy the equation.
     
  10. Aug 23, 2010 #9
    Would there be a reasonably simple way to get to the two results given without substituting them into the first equation? I have a feeling that I'm supposed to do it that way, and not just by replacing x with the answers given.

    It seems like I could get to that form with some rearranging and factorising but I'm not sure where to begin with that, or if there's any specific techniques that might help.
     
  11. Aug 23, 2010 #10
    Go back to your original equation and rewrite in the form f(x) = 0. Observe that you can factor out ax, so your equation looks like axg(x) = 0. Now g(x) is a cubic polynomial and you know that 1 is a root. Use long division to compute g(x)/(x-1). The result is a polynomial of degree 2. Now use the quadratic formula to find its roots. You end up with the result in your first post.
     
  12. Aug 23, 2010 #11
    Thanks, that worked perfectly!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Solving a cubic equation
  1. Cubic Equation (Replies: 2)

Loading...