Question

How do you find all the complex roots of `5x^6 + x^4 -3 = 0`?

Answer 1

Solve as a cubic equation in `1/x^2` using Cardano's method.

Explanation:

Let `t = 1/x^2`

Then our polynomial equation becomes: `5/t^3+1/t^2-3 = 0`

Multiply through by `t^3` and rearrange to get:

`3t^3-t-5 = 0`

Use Cardano's method:

Let `t = u+v`

Then:

`0 = 3(u+v)^3-(u+v)-5`

`=3u^3+3v^3+(9uv-1)(u+v)-5`

Add the constraint `v = 1/(9u)` (causing `(9uv-1) = 0`) to get:

`=3u^3+1/(243u^3)-5`

Multiply through by `243u^3` to get a quadratic equation in `u^3`:

`729(u^3)^2-1215(u^3)+1 = 0`

This has roots given by the quadratic formula:

`u^3 = (1215+-sqrt(1215^2-(4*729)))/1458`

`=(1215+-sqrt(1476225-2916))/1458`

`=(1215+-sqrt(1473309))/1458`

`=(1215+-27 sqrt(2021))/1458`

`=(45+-sqrt(2021))/54`

The derivation was symmetric in `u` and `v`, so one of these roots is `u^3` and the other `v^3`. Hence the Real root of our cubic in `t` is:

`t_1 = root(3)((45+sqrt(2021))/54) + root(3)((45-sqrt(2021))/54)`

`= 1/3 root(3)((45+sqrt(2021))/2) + 1/3 root(3)((45-sqrt(2021))/2)`

and the non-Real Complex roots are given by:

`t_2 = 1/3 omega root(3)((45+sqrt(2021))/2) + 1/3 omega^2 root(3)((45-sqrt(2021))/2)`

`t_3 = 1/3 omega^2 root(3)((45+sqrt(2021))/2) + 1/3 omega root(3)((45-sqrt(2021))/2)`

where `omega = -1/2 + sqrt(3)/2 i` is the primitive Complex cube root of `1`

Hence the roots of our original equation are:

`x_(1,2) = +-1/sqrt(t_1)`

`x_(3,4) = +-1/sqrt(t_2)`

`x_(5,6) = +-1/sqrt(t_3)`

For example,

`x_3 = 1/sqrt(1/3 omega root(3)((45+sqrt(2021))/2) + 1/3 omega^2 root(3)((45-sqrt(2021))/2)`

I have previously explained how to get the square root of a Complex number in `a+bi` form, which you can use to help get these roots into `a+bi` form.

See: http://socratic.org/questions/how-do-you-find-the-square-root-of-an-imaginary-number-of-the-form-a-bi