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

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.