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Solving uartics

  1. Sep 23, 2012 #1
    Is there a method to finding the roots of quartics besides Ferrari's formula?

    I have the equation

    [tex]x^{4}+5x^{2}+4x+5=0[/tex]

    I know one of the factor is something like $$x^{2}+x+1$$ and the other one can be found using sythetic division, but how can I find the factors without knowing one of them in the first place?

    Thanks.

    My bad about the title. It's supposed to be Quartics instead of uartics.
     
  2. jcsd
  3. Sep 23, 2012 #2

    Mentallic

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    If the quartic has "nice" quadratic factors, then they will be of the form:

    [tex]x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)[/tex]

    Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the [itex]\pm[/itex] operator.
     
  4. Sep 23, 2012 #3
    Oh, I see, but how do we know when the factors are "nice"?
     
  5. Sep 23, 2012 #4

    Mentallic

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    We don't. We can always check to see if they are, just like we check to see if there are any rational roots.

    With the quartic [itex]x^4+5x^2+4x+5[/itex] we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, [itex]x^4+6x^2+4x+5[/itex] then we find no solutions for a,b in the system of equations.
     
  6. Sep 23, 2012 #5
    Ok, thanks!
     
  7. Sep 23, 2012 #6

    mathwonk

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  8. Sep 24, 2012 #7
    Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link :smile:
     
  9. Sep 24, 2012 #8
    I end up getting two systems but one of them has no solution. Did I proceed correctly?
     
  10. Sep 24, 2012 #9

    Mentallic

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    I don't know, what did you get?
     
  11. Sep 24, 2012 #10
    Sorry for not having LATEX since I'm on my phone, but I got:

    System 1:

    a+b=0
    5+1+ab=5
    5a+b=4

    System 2:

    a+b=0
    -5-1+ab=5
    -5a-b=4
     
  12. Sep 24, 2012 #11

    Mentallic

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    Right, so simplifying this system, we have

    a+b=0
    ab=-1
    5a+b=4

    Which we can then deduce,
    a=-b
    therefore,
    a2=1 -> a=[itex]\pm1[/itex]
    4a=4 -> a=1, b=-1



    For this system we have no real solution, so what does that tell you?
     
  13. Sep 24, 2012 #12
    For the second system, I don't get any solutions. Therefore, that can't be the right system.
     
  14. Sep 24, 2012 #13

    Mentallic

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    Exactly, so what must your factors be?
     
  15. Sep 25, 2012 #14
    Ah, that's what I thought. Thanks!
     
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