Ferrari-Cardano derivation of the quartic formula

Given a quartic equation $x^4+a{x}^3+b{x}^2+cx+d=0$, apply the Tchirnhaus transformation $x\mapsto y-\frac{a}{4}$ to obtain

$$y^4+p{y}^2+qy+r=0$$ (1)

where

$p$ $=$ $b-{\displaystyle \frac{3a^2}{8}}$

$q$ $=$ $c-{\displaystyle \frac{ab}{2}}+{\displaystyle \frac{a^3}{8}}$

$r$ $=$ $d-{\displaystyle \frac{ac}{4}}+{\displaystyle \frac{a^{2}b}{16}}-{\displaystyle \frac{3a^4}{256}}$

Clearly a solution to Equation (1) solves the original, so we replace the original equation with Equation (1). Move $qy+r$ to the other side and complete the square on the left to get:

$${(y^2+p)}^2=p{y}^2-qy+(p^2-r).$$

We now wish to add the quantity ${(y^2+p+z)}^2-{(y^2+p)}^2$ to both sides, for some unspecified value of $z$ whose purpose will be made clear in what follows. Note that ${(y^2+p+z)}^2-{(y^2+p)}^2$ is a quadratic in $y$. Carrying out this addition, we get

$${(y^2+p+z)}^2=(p+2z)y^2-qy+(z^2+2pz+p^2-r)$$ (2)

The goal is now to choose a value for $z$ which makes the right hand side of Equation (2) a perfect square. The right hand side is a quadratic polynomial in $y$ whose discriminant is

$$-8z^3-20p{z}^2+(8r-16p^2)z+q^2+4pr-4p^3.$$

Our goal will be achieved if we can find a value for $z$ which makes this discriminant zero. But the above polynomial is a cubic polynomial in $z$, so its roots can be found using the cubic formula. Choosing then such a value for $z$, we may rewrite Equation (2) as

$${(y^2+p+z)}^2={(sy+t)}^2$$

for some (complicated!) values $s$ and $t$, and then taking the square root of both sides and solving the resulting quadratic equation in $y$ provides a root of Equation (1).

Title	Ferrari-Cardano derivation of the quartic formula
Canonical name	FerrariCardanoDerivationOfTheQuarticFormula
Date of creation	2013-03-22 12:37:21
Last modified on	2013-03-22 12:37:21
Owner	djao (24)
Last modified by	djao (24)
Numerical id	8
Author	djao (24)
Entry type	Proof
Classification	msc 12D10
Related topic	CardanosDerivationOfTheCubicFormula