A Modern Vision of the Work of Cardano and Ferrari on Quartics - An Eulerian Idea About Biquadratics

Can we find the solutions of x⁴ + x² - 1 = 0? Defining y = x², we get y² + y - 1 = 0, whose solutions are -1/2 ± √ 5/2. Then the four solutions of the original equation stem from x² = -1/2 ± √ 5/2; they are $$\pm\sqrt{-1+\sqrt{5}\over 2}, \quad \pm i\sqrt{1+\sqrt{5}\over 2}.$$ Problems of this sort were discussed in section 2. It is clear that, given the equation x⁴ + ax² + b = 0, with real coefficients, the transformation y = x² works well whenever a² - 4b > 0. We use the phrase "works well" to mean that we do not have to worry about finding the square root of a complex number. From a slightly different perspective we might observe that x⁴ + ax² = -b implies (x²+a/2)² = a²/4 - b; thus, (x² + a/2)² = (√(a²-4b)/2)², an equality that allows us to conclude that x² + a/2=(1/2)√(a² - 4b) or x² + a/2 = -(1/2)√(a²-4b). From these quadratic equations the four solutions (real or complex) do follow.

At the beginning of section 4 we analyzed the equation x⁴ + x² + 1 = 0 after writing x⁴ + 1 = -x² and "completing squares" in the sense that (x² + 1)² = -x² + 2x². Is it possible to extend this idea to all biquadratics whenever the transformation cannot be applied with ease? In his quest of a proof of the Fundamental Theorem of Algebra, Leonhard Euler (1707-1783) considered the equation x⁴ + ax² + b = 0, where a² - 4b < 0 ([5], p. 364). The inequality a² < 4b implies a <= |a| < 2√b, hence 2√b - a > 0. We have x⁴ + b = -ax², which is equivalent to (x² + √b)² = -ax² + 2√bx². Therefore (x² + √b)² = (x√(2√b-a)², which in turn leads to x² + √b = x√(2√b-a) or x² + √b = -x√(2√b-a). The two quadratics provide the four solutions, real or complex, of x⁴ + ax² + b = 0. We note that the expression a² - 4b, called the "discriminant" of the biquadratic, determines the path to be followed. We have succeeded in solving the most general biquadratic with real coefficients.

To illustrate the procedure let us analyze with care the equation x⁴ + 12 = 6x², one of Cardano's original equations that we mentioned in section 2. We note that its discriminant is negative, thus we write (x²+√12)² = 6x² + 2(√12)x², that is to say (x² + √12)² = (√(6 + 2√12)x)². Hence x² + √12 = √(6 + 2√12)x or x² + √12 = -√(6+2√12)x. The first equation has the two complex solutions $${1\over 2}\sqrt{6+2\sqrt{12}}\pm i{1\over 2}\sqrt{2\sqrt{12}-6}$$ while the second equation has the two complex solutions $$-{1\over 2}\sqrt{6+2\sqrt{12}} \pm i{1\over 2}\sqrt{2\sqrt{12}-6}.$$ These are the four solutions of the given biquadratic in the complex field. Cardano was right in claiming that x⁴ + 12 = 6x² has no solutions in the "real" realm, the only type acceptable to him and his contemporaries.