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# A Modern Vision of the Work of Cardano and Ferrari on Quartics

## The Most General Setting

After having discussed several examples of quartics, and the most general biquadratic, it is time to look at Ferrari's method from a modern perspective. Let us consider the equation x4 + d1x3 + d2x2 + d3x + d4 = 0, where the coefficients are real numbers. Using the transformation x=y - d1/4 we can convert it into an equation of the type y4 + ay2 + by + c = 0. We may assume that b ¹ 0 because b = 0 leads to a biquadratic, whose method of solution was analyzed in the previous section.

Next we will apply Ferrari's technique. Indeed y4 = -ay2 - by - c. So (y2 + z)2 = 2y2z + z2 - ay2 - by - c for any z. Our goal is to find a real number z such that the right hand side of the last equality becomes a perfect square. Rearranging terms we get $$(y^2+z)^2 = (2z-a)y^2-by+(z^2-c)\ \ \ \ (2).$$

The expression (2z - a)y2 - by + (z2 - c) will be a perfect square provided its discriminant is zero, i.e. b2 - 4(2z - a)(z2 - c) = 0 or, what is the same, 8z3 - 4az2 - 8cz + (4ac - b2) = 0. This is a cubic equation, customarily called a "resolvent cubic". Let z1 be a real solution of it. Evidently 2z1 - a ¹ 0 because otherwise the resolvent cubic implies b2 = 0, which in turn leads to b = 0. Then $$(2z_1-a)y^2-by+(z_1^2-c) = (2z_1-a)\left(y-{b\over 2(2z_1-a)}\right)^2.$$ Thus, thanks to (2) we get $$(y^2+z_1)^2 = (2z_1-a)\left(y-{b\over 2(2z_1-a)}\right)^2.$$ Consequently, $$y^2+z_1 = \sqrt{2z_1-a}\left(y-{b\over 2(2z_1-a)}\right) \quad {\rm or} \quad y^2+z_1 = -\sqrt{2z_1-a}\left(y-{b\over 2(2z_1-a)}\right).$$

Solving these quadratics we get the four solutions of y4 + ay2 + by + c = 0, which in turn lead to the four solutions of the original quartic through the use of the transformation x = y - d1/4. We have to keep in mind that if 2z1 - a < 0 we will have to deal with a quadratic that happens to have complex coefficients.

An alternative way to solve y4 + ay2 + by + c = 0, through Ferrari's technique, starts by noticing that (y2 + a/2)2 = -by -c + a2/4 (recall Cardano's solution of problem VIII, discussed in detail in section 6). For any z we will have (y2 + a/2 + z)2 = -by - c + a2/4 + z2 + 2z(y2 + a/2), that is to say (y2 + a/2 + z)2 = 2zy2 - by + (z2- c + a2/4 + az). In order to have a perfect square to the right of this expression we need to make the discriminant equal to zero, thus b2 - 8z(z2 - c + a2/4 + az) = 0. In other words, we need to solve the equation 8z3 + 8az2 - 8cz + 2a2z - b2 = 0, a resolvent cubic different from the one we found before.

Let z1 be a real solution of it, which has to be different from zero because otherwise b = 0; therefore (y2 + a/2 + z1)2 = 2z1(y - b/4z1)2. The four solutions of y4 + ay2 + by + c = 0 will stem from the quadratics (y2 + a/2 + z1) = √(2z1)(y - b/4z1), (y2 + a/2 + z1) = -(2z1)(y b/4z1). Explicitly, the four solutions (real or complex) are $$y={1\over 2}\left(\sqrt{2z_1} \pm \sqrt{-2z_1-2a-b\sqrt{2/z_1}}\right),\quad y = {1\over 2}\left(-\sqrt{2z_1}\pm \sqrt{-2z_1-2a-b\sqrt{2/z_1}}\right).$$

Ferrari's technique works for all quartics, although it is not of practical use in the case of biquadratics. For instance, let us analyze the biquadratic equation x4 + 12 = 6x2, whose complex solutions were found at the end of the previous section. We have (x2+z)2 = 6x2 - 12 + z2 + 2zx2 for any z. That is to say $$(x^2+z)^2 = (2z+6)x^2+(z^2-12)\ \ \ \ (3).$$

We wish to find z such that the right hand side of (3) is a perfect square. This will happen if its discriminant is zero; thus, -4(2z + 6)(z2 - 12)=0. We can choose the solution √12. Replacing this value in (3) we get (x2 + 2√3)2 = (x√(4√3+6))2. The four solutions follow from this equation; a lot of work taking into consideration that there is a straightforward alternative for biquadratics, which we analyzed in sections 2 and 8.

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Helfgott, Harald and Michel Helfgott, "A Modern Vision of the Work of Cardano and Ferrari on Quartics," Loci (June 2009), DOI: 10.4169/loci003312