How do I solve a sextic?

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.

There is, in general, no analytic solution for sextic equations (this has been proven). For this one in particular, we have... $$x^{6} + x^{2} - 9 = 0$$ Let $u = x^{2}$, giving us...$$u^{3} + u - 9$$ Now approach it as a cubic.

LCKurtz
Homework Helper
Gold Member
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.

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HallsofIvy
Homework Helper
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.
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.

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.
LCKurtz, I accidently clicked on "edit" when I intended "quote". Sorry about that. I think I put your post back together correctly.

LCKurtz
Homework Helper
Gold Member
While it is possible to solve cubics such as this by hand, you likely have better things to do...

Well, I don't!

Heh. Nothing else to do eh?
LCKurtz, I accidently clicked on "edit" when I intended "quote". Sorry about that. I think I put your post back together correctly.

"With great power comes great responsibility." -- Spiderman

Heh. Nothing else to do eh?

"With great power comes great responsibility." -- Spiderman

Pfft...he was only quoting his Aunt. Your knowledge of Marvel comics is insufficient for you to be giving advice on this forum.