Solve the equation $x^{4} + 4 x^{3} + 8 = 0$ $x^4+4x^3+8=0$ ?

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Solve the equation $x^{4} + 4 x^{3} + 8 = 0$ $x^4+4x^3+8=0$ ?

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

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Answer 1 · 2017-03-24T11:02:59

See below.

Explanation:

Consider the polynomial

$x^{4} + 2 a x^{3} + b = 0$ $x^4+2ax^3+b=0$

and also the expansion

${(x^{2} + a x + ξ)}^{2} = x^{4} + 2 a x^{3} + (a^{2} + 2 ξ) x^{2} + 2 a ξ x + ξ^{2}$ $(x^2+ax+xi)^2=x^4+2ax^3+(a^2+2xi)x^2+2axix+xi^2$

(here $ξ$ $xi$ is a dummy parameter)

now substituting $x^{4} + 2 a x^{3} = - b$ $x^4+2ax^3 = -b$ we obtain

${(x^{2} + a x + ξ)}^{2} = (a^{2} + 2 ξ) x^{2} + 2 a ξ x + ξ^{2} - b$ $(x^2+ax+xi)^2=(a^2+2xi)x^2+2axix+xi^2-b$

and now choosing properly $ξ$ $xi$ to make $(a^{2} + 2 ξ) x^{2} + 2 a ξ x + ξ^{2} - b$ $(a^2+2xi)x^2+2axix+xi^2-b$ a square with

$2 ξ^{3} - 2 b ξ - a^{2} b = 0$ $2 xi^3- 2 b xi -a^2 b=0$

or

$ξ^{3} - 8 ξ - 16 = 0$ $xi^3-8xi-16=0$

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Note that solving $(a^{2} + 2 ξ) x^{2} + 2 a ξ x + ξ^{2} - b = 0$ $(a^2+2xi)x^2+2axix+xi^2-b=0$ or

$x = \frac{- a ξ \pm \sqrt{a^{2} b + 2 b ξ - 2 ξ^{3}}}{a^{2} + 2 ξ}$ $x=(-a xi pm sqrt[a^2 b + 2 b xi - 2 xi^3])/(a^2 + 2 xi)$ if we choose
$ξ_{0}$ $xi_0$ such that $a^{2} b + 2 b ξ_{0} - 2 ξ_{0}^{3} = 0$ $a^2 b + 2 b xi_0 - 2 xi_0^3=0$ then

$(a^{2} + 2 ξ_{0}) x^{2} + 2 a ξ_{0} x + ξ_{0}^{2} - b = (a^{2} + 2 ξ_{0}) {(x + a ξ_{0})}_{0}^{2}$ $(a^2+2xi_0)x^2+2axi_0x+xi_0^2-b = (a^2+2xi_0)(x+a xi_0)^2$
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Now solving for $ξ$ $xi$

$ξ = ξ_{0} = \frac{1}{3} {(216 - 24 \sqrt{57})}^{\frac{1}{3}} + \frac{2 {(9 + \sqrt{57})}^{\frac{1}{3}}}{3^{\frac{2}{3}}}$ $xi=xi_0=1/3 (216 - 24 sqrt[57])^(1/3) + (2 (9 + sqrt[57])^(1/3))/3^(2/3)$ which is the only real root.

Now

$(a^{2} + 2 ξ_{0}) x^{2} + 2 a ξ_{0} x + ξ_{0}^{2} - b = (a^{2} + 2 ξ_{0}) {(x + a ξ_{0})}_{0}^{2}$ $(a^2+2xi_0)x^2+2axi_0x+xi_0^2-b = (a^2+2xi_0)(x+ axi_0)^2$

and the former quartic remains

${(x^{2} + a x + ξ_{0})}_{0}^{2} = (a^{2} + 2 ξ_{0}) {(x + a ξ_{0})}_{0}^{2}$ $(x^2+ax+xi_0)^2=(a^2+2xi_0)(x+ axi_0)^2$ or

${(x^{2} + a x + ξ_{0})}_{0}^{2} - (a^{2} + 2 ξ_{0}) {(x + a ξ_{0})}_{0}^{2} = 0$ $(x^2+ax+xi_0)^2-(a^2+2xi_0)(x+ axi_0)^2=0$ or

$(x^{2} + a x + ξ_{0} + \sqrt{a^{2} + 2 ξ_{0}} (x + a ξ_{0})) (x^{2} + a x + ξ_{0} - \sqrt{a^{2} + 2 ξ_{0}} (x + a ξ_{0})) = 0$ $(x^2+ax+xi_0+sqrt(a^2+2xi_0)(x+axi_0))(x^2+ax+xi_0-sqrt(a^2+2xi_0)(x+axi_0))=0$

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.

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Solve the equation $x^{4} + 4 x^{3} + 8 = 0$ $x^4+4x^3+8=0$ ?

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

1 Answer

Explanation:

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Solve the equation x^4+4x^3+8=0x4+4x3+8=0x^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=0x4+4x3+8=0x^4+4x^3+8=0. Are the cubics the same? Further, Use either method to obtain the roots of this equation.

1 Answer

Explanation:

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Solve the equation $x^{4} + 4 x^{3} + 8 = 0$ $x^4+4x^3+8=0$ ?

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