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单词 FerrariCardanoDerivationOfTheQuarticFormula
释义

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)

where

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:

(y2+p)2=py2-qy+(p2-r).

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

-8z3-20pz2+(8r-16p2)z+q2+4pr-4p3.

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

(y2+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).

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更新时间:2025/5/4 12:11:26