# A Modern Vision of the Work of Cardano and Ferrari on Quartics - The Most General Setting

Author(s):
Harald Helfgott (University of Bristol) and Michel Helfgott (East Tennessee State University)

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.