# How do I solve a sextic?

1. Oct 23, 2011

### minusequals

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. Oct 23, 2011

### Number Nine

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.

3. Oct 23, 2011

### LCKurtz

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
4. Oct 24, 2011

### HallsofIvy

Staff Emeritus
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.

5. Oct 24, 2011

### LCKurtz

Heh. Nothing else to do eh?
"With great power comes great responsibility." -- Spiderman

6. Oct 24, 2011

### Number Nine

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