Ferrari-Cardano derivation of the quartic formula

Given a quartic equationMathworldPlanetmath x4+ax3+bx2+cx+d=0, apply the Tchirnhaus transformation xy-a4 to obtain

y4+py2+qy+r=0 (1)


p = b-3a28
q = c-ab2+a38
r = d-ac4+a2b16-3a4256

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 completePlanetmathPlanetmathPlanetmathPlanetmath the square on the left to get:


We now wish to add the quantity (y2+p+z)2-(y2+p)2 to both sides, for some unspecified value of z whose purpose will be made clear in what follows. Note that (y2+p+z)2-(y2+p)2 is a quadratic in y. Carrying out this addition, we get

(y2+p+z)2=(p+2z)y2-qy+(z2+2pz+p2-r) (2)

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


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


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