Solve the equation #x^4+4x^3+8=0#?

Obtain the resolvent cubics, by Descartes' method and by Ferrari's method, of the equation #x^4+4x^3+8=0#. Are the cubics the same? Further, Use either method to obtain the roots of this equation.

1 Answer
Mar 24, 2017

Answer:

See below.

Explanation:

Consider the polynomial

#x^4+2ax^3+b=0#

and also the expansion

#(x^2+ax+xi)^2=x^4+2ax^3+(a^2+2xi)x^2+2axix+xi^2#

(here #xi# is a dummy parameter)

now substituting #x^4+2ax^3 = -b# we obtain

#(x^2+ax+xi)^2=(a^2+2xi)x^2+2axix+xi^2-b#

and now choosing properly #xi# to make #(a^2+2xi)x^2+2axix+xi^2-b# a square with

# 2 xi^3- 2 b xi -a^2 b=0#

or

#xi^3-8xi-16=0#

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Note that solving #(a^2+2xi)x^2+2axix+xi^2-b=0# or

#x=(-a xi pm sqrt[a^2 b + 2 b xi - 2 xi^3])/(a^2 + 2 xi)# if we choose
#xi_0# such that #a^2 b + 2 b xi_0 - 2 xi_0^3=0# then

#(a^2+2xi_0)x^2+2axi_0x+xi_0^2-b = (a^2+2xi_0)(x+a xi_0)^2#
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Now solving for #xi#

#xi=xi_0=1/3 (216 - 24 sqrt[57])^(1/3) + (2 (9 + sqrt[57])^(1/3))/3^(2/3)# which is the only real root.

Now

#(a^2+2xi_0)x^2+2axi_0x+xi_0^2-b = (a^2+2xi_0)(x+ axi_0)^2#

and the former quartic remains

#(x^2+ax+xi_0)^2=(a^2+2xi_0)(x+ axi_0)^2# or

#(x^2+ax+xi_0)^2-(a^2+2xi_0)(x+ axi_0)^2=0# or

#(x^2+ax+xi_0+sqrt(a^2+2xi_0)(x+axi_0))(x^2+ax+xi_0-sqrt(a^2+2xi_0)(x+axi_0))=0#

resulting in the resolution of

#{(x^2+(a+sqrt(a^2+2xi_0))x+xi_0+axi_0sqrt(a^2+2xi_0)=0),(x^2+(a-sqrt(a^2+2xi_0))x+xi_0-axi_0sqrt(a^2+2xi_0)=0):}#

This is left as an exercise for the proficient reader.

NOTE: This is the so called Ferrari's method.