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

1 Answer
Feb 12, 2016

Answer:

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