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How do I solve a sextic?

  1. Oct 23, 2011 #1
    where do a cubic and a circle intersect? y=x^3, x^2+y^2=9? Descartes' rule of signs and rational zero theorem at least tell me we have a symmetical +- irrational root; what tools do I use next? is sin theta = cos^3 theta going anywhere? x^2 + x^6 - 9 = 0. I have seen a graphical solution, but have no idea which analytical tools to use.
     
  2. jcsd
  3. Oct 23, 2011 #2
    There is, in general, no analytic solution for sextic equations (this has been proven). For this one in particular, we have... [tex]x^{6} + x^{2} - 9 = 0[/tex] Let [itex]u = x^{2}[/itex], giving us...[tex]u^{3} + u - 9[/tex] Now approach it as a cubic.
     
  4. Oct 23, 2011 #3

    LCKurtz

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    By letting u = x2 you get a cubic in u. While it is possible to solve cubics such as this by hand, you likely have better things to do. You could solve it numerically or use a program such as Maple to write out an explicit formula. Here's the exact answer given by Maple for the positive value of x where they intersect (hard to read and parse):

    (1/6)*sqrt(6)*sqrt((972+12*sqrt(6573))^(1/3)*((972+12*sqrt(6573))^(2/3)-12))/(972+12*sqrt(6573))^(1/3)

    or approximately 1.385703836.
     
    Last edited by a moderator: Oct 24, 2011
  5. Oct 24, 2011 #4

    HallsofIvy

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    Well, I don't! Using the cubic formula (and, I will admit, a calculator) I got 1.9195 as a solution to the cubic, to four decimal places. The square root of that is 1.3855, basically your answer. Using that you can reduce the problem to a quadratic.

    LCKurtz, I accidently clicked on "edit" when I intended "quote". Sorry about that. I think I put your post back together correctly.
     
  6. Oct 24, 2011 #5

    LCKurtz

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    Heh. Nothing else to do eh? :cool:
    "With great power comes great responsibility." -- Spiderman
     
  7. Oct 24, 2011 #6
    Pfft...he was only quoting his Aunt. Your knowledge of Marvel comics is insufficient for you to be giving advice on this forum.
     
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