# Ferrari-Cardano derivation of the quartic formula

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

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

where

 $\displaystyle p$ $\displaystyle=$ $\displaystyle b-\frac{3a^{2}}{8}$ $\displaystyle q$ $\displaystyle=$ $\displaystyle c-\frac{ab}{2}+\frac{a^{3}}{8}$ $\displaystyle r$ $\displaystyle=$ $\displaystyle d-\frac{ac}{4}+\frac{a^{2}b}{16}-\frac{3a^{4}}{256}$

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:

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

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

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

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 FerrariCardanoDerivationOfTheQuarticFormula 2013-03-22 12:37:21 2013-03-22 12:37:21 djao (24) djao (24) 8 djao (24) Proof msc 12D10 CardanosDerivationOfTheCubicFormula