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.