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At the very beginning of *Ars Magna*, five polynomial equations of the fourth order appear ([1], pp. 10-11), namely *x*^{4 }= 81, *x*^{4 }+ 3*x*^{2 }= 28, *x*^{4 }+ 12 = 7*x*^{2}, *x*^{4 }+ 12 = 6*x*^{2}, and *x*^{4 }= 2*x*^{2 }+ 8. Cardano ascertains that 3 and -3 are solutions of *x*^{4 }= 81, 2 and -2 are solutions of *x*^{4 }+ 3*x*^{2 }= 28, while 2, -2, √3, -√3 are solutions of *x*^{4 }+ 12 = 7*x*^{2}. Moreover, Cardano writes that *x*^{4 }+ 12 = 6*x*^{2} has no solutions and *x*^{4 }= 2*x*^{2 }+ 8 has 2, -2 as solutions. It is to be noted that Cardano calls positive solutions "true" while negative solutions are called "fictitious". No mention is made of complex solutions of equations since complex numbers were not accepted during Cardano's time.

We can check that Cardano has found all the solutions in the real field. For instance, by long division we get *x*^{4 }+ 3*x*^{2 }- 28 = (*x*-2)(*x*+2)(*x*^{2}+7), so 2 and -2 are the only real solutions of *x*^{4 }+ 3*x*^{2 }= 28 . Later, in chapter 24 of *Ars Magna*, we find the method used by Cardano. He starts with the biquadratic equation *x*^{4 }+ 2*x*^{2 }= 10 and claims that it is enough to find the solutions of *x*^{2 }+ 2*x *= 10 and then take the square roots of them. The corresponding quadratic has only one positive solution, namely \( \sqrt{11} - 1 \); thus, \( \pm \sqrt{ \sqrt{11} - 1} \) are the only "real" solutions of the biquadratic. Although \( -\sqrt{11} - 1 \) is also a real solution of *x*^{2 }+ 2*x *= 10, Cardano does not consider \( \pm i\sqrt{ \sqrt{11} + 1} \) as solutions of the given biquadratic since they are complex numbers.

From a modern perspective we can see that Cardano noticed that to solve *x*^{4 }+ 2*x*^{2}=10 it is enough to define the transformation *y *= *x*^{2} and then analyze the equation *y*^{2 }+ 2*y *= 10. However, we run into difficulties when we try to solve *x*^{4 }+ 12 = 6*x*^{2}, one of the equations that appear at the beginning of *Ars Magna*, because the transformation *y *= *x*^{2} leads to *y*^{2 }- 6*y *+ 12 = 0, whose solutions are 3 ± *i*√3. Thus, we would have to solve *x*^{2 }= 3 ± *i*√3. This is a quadratic equation with complex coefficients; it is no wonder that Cardano claims that *x*^{4 }+ 12 = 6*x*^{2} has no solutions - obviously he is thinking about solutions in the real field. In section 8 we will analyze this equation in detail.

Problem I, chapter 39 of *Ars Magna*, leads to a biquadratic: "Find three proportional quantities of which the square of the first is equal to the second and third, and the square of the third is equal to the squares of the second and first." Let the quantities be *y*, *yx*, *yx*^{2}, where *x* is the constant of proportionality. The problem demands that *y*^{2 }= *yx *+ *yx*^{2} and (*yx*^{2})^{2 }= *y*^{2 }+ (*yx*)^{2}. The first equation leads to *y *= *x *+ *x*^{2} while the second equation leads to *x*^{4 }= *x*^{2 }+ 1. Defining *z *= *x*^{2} we get *z*^{2 }= *z *+ 1, whose solutions are $$z={1\over 2} \pm {\sqrt{5}\over 2}.$$ We are not interested in the negative solution since it will lead to the complex solutions $$x=\pm i\sqrt{{\sqrt{5}\over 2}-{1\over 2}}.$$ Rather we deal with *x*^{2 }= *z *= 1/2+√5/2, whose positive solution is $$x=\sqrt{{1\over 2}+{\sqrt{5}\over 2}}.$$ Then $$y=\sqrt{{1\over 2}+{\sqrt{5}\over 2}} + {1\over 2} +{\sqrt{5}\over 2},$$ and we can conclude that the three quantities are $$\sqrt{{1\over 2}+{\sqrt{5}\over 2}} + {1\over 2}+{\sqrt{5}\over 2},\quad \left(\sqrt{{1\over 2}+{\sqrt{5}\over 2}} + {1\over 2}+{\sqrt{5}\over 2}\right)\sqrt{{1\over 2}+{\sqrt{5}\over 2}}\quad {\rm and} \quad \left(\sqrt{{1\over 2}+{\sqrt{5}\over 2}} + {1\over 2}+{\sqrt{5}\over 2}\right)\left({1\over 2}+{\sqrt{5}\over 2}\right).$$

Of a similar nature is problem X, chapter 39 of *Ars Magna*: "Find three proportional numbers the sum of which is 8 and the square of the third of which is equal to the sum of the squares of the first and second". Let the numbers be *y*, *yx*, and *yx*^{2}, where *x* is the constant of proportionality. Then *y*^{2}*x*^{4 }= *y*^{2 }+ *y*^{2}*x*^{2}, which leads to *x*^{4 }= *x*^{2}+1. The transformation *z *= *x*^{2} provides us with only one positive solution, namely $$x=\sqrt{1+\sqrt{5}\over 2}.$$ But *y *+ *yx *+ *yx*^{2 }= 8, so $$y={8\over \left(1+\sqrt{1+\sqrt{5}\over 2}+ {1+\sqrt{5}\over 2}\right)}.$$ The problem has come to an end because we have been able to find the values of *y* and *x.*

It is to be noted that Cardano reached the solution to problem X following a somewhat different procedure. He denotes the three proportional numbers 1, *x*, *x*^{2}. Since(*x*^{2})^{2 }= *x*^{2 }+ 1 we will have *x*^{4 }= *x*^{2}+1, whose only positive solution is $$x=\sqrt{1+\sqrt{5}\over 2}.$$ But $$r =1+\sqrt{1+\sqrt{5}\over 2}+ {1+\sqrt{5}\over 2}$$ would have to be 8, which it is not. To remedy this situation Cardano uses the rule of three and concludes that $$y={8\over \left(1+\sqrt{1+\sqrt{5}\over 2}+ {1+\sqrt{5}\over 2}\right)}$$ is the first quantity, \( y\sqrt{\frac{1+\sqrt{5}}{2}} \) is the second quantity, and \( y\left(\frac{1+\sqrt{5}}{2}\right) \) is the third quantity.

Harald Helfgott (University of Bristol) and Michel Helfgott (East Tennessee State University), "A Modern Vision of the Work of Cardano and Ferrari on Quartics - Biquadratic Equations," *Loci* (February 2010), DOI:10.4169/loci003312