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

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

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.