Let us try to solve the biquadratic *x*^{4 }+ *x*^{2} + 1 = 0, an equation that cannot be analyzed with ease if we were to use the transformation *y* = *x*^{2} because we would have to calculate the square root of a complex number. Instead, "completing squares" we get (*x*^{2}+ 1)^{2} = *x*^{2}; thus, *x*^{2} + 1 = *x* or *x*^{2} + 1 = -*x*. The first quadratic has the solutions 1/2 ±* i*√3/2 while the second has the solutions -1/2 ± *i*√3/2; these four numbers are precisely the solutions of the given biquadratic. The crux of the matter was to reach the equation (*x*^{2} + 1)^{2} = *x*^{2}, wherein two perfect squares appear at each side of the equality sign. Precisely this idea will help us to better understand what comes next.

Lodovico Ferrari invented a systematic procedure to solve all quartics and he is duly recognized in *Ars Magna *for this important achievement ([1] p. 237). As we will see, the procedure leads to the solution of a cubic, called the "Resolvent Cubic". A solution of the latter can be found by inspection - if we are lucky - or we have to use the Cardano-Tartaglia technique for solving cubics (Niccolo Tartaglia (1500-1557) gave Cardano, in 1539, the method to solve *y*^{3}+ *py* = *q*, where *p*,*q* > 0, and two other particular cases of the cubic). To illustrate Ferrari's method let us discuss in detail two problems from chapter 39 of *Ars Magna*.

Figure 3: Problem XII from chapter 39

Let us start with problem XII: Solve the equation *x*^{4} + 3 = 12*x*, which is equivalent to (*x*^{2})^{2} = 12*x *- 3. For every *b*, this implies (*x*^{2} + *b*)^{2} = 2*bx*^{2} + 12*x* + (*b*^{2} - 3). Now, in general, a quadratic polynomial is the square of another polynomial if the discriminant is equal to 0. Thus, we must find *b* such that the discriminant 12^{2} - 4(2*b*)(*b*^{2 }- 3) of the polynomial 2*bx*^{2 }+ 12*x* + *b*^{2 }- 3 equals 0. Expanding, we see that this is the same as asking that *b*^{3 }= 3*b* + 18. By inspection we find that *b* = 3 is a solution. Using this value of *b* leads us to (*x*^{2} + 3)^{2} = 2*3*x*^{2} + 12*x *+ (3^{2} - 3), i.e. (*x*^{2} + 3)^{2} = 6(*x*^{2} + 2*x* + 1), which is equivalent to (*x*^{2} + 3)^{2} = (√6(*x* + 1))^{2}. Therefore *x*^{2}+ 3 = √6(*x* + 1) or *x*^{2} + 3 = -√6(*x* + 1). The first quadratic equation provides the positive real solutions √6/2 ± (1/2)√(-6 + 4√6) and the second quadratic equation provides the conjugate pair -√6/2 ± (*i*/2)√(6-4√6). These are precisely the solutions of the original quartic, although Cardano considers only √6/2 + (1/2)√(-6 + 4√6).

It should be noted that Cardano does not use the term "discriminant"; given any quadratic polynomial he realizes that it will have a double root provided that "the square of half the middle quantity, *x*, equals the product of the extremes" ([1], p. 244). In modern symbolism we would say that any quadratic polynomial *px*^{2} + *qx* + *r* will have a double root if (*q*/2)^{2} = *pr*, which is equivalent to demand that *q*^{2} - 4*pr* = 0; thus, the discriminant of the quadratic polynomial has to be zero.

If we apply the preceding method to *x*^{4} + 4*x* + 8 = 10*x*^{2} (problem IX) we would get, for any *z*, the equality (*x*^{2} + *z*)^{2} = 10*x*^{2} - 4*x *- 8 + 2*zx*^{2} + *z*^{2}, that is to say (*x*^{2} + *z*)^{2} = (10 + 2*z*)*x*^{2} - 4*x* + (*z*^{2 }- 8). We wish to find a real value of *z* that will make the expression to the right of the equal sign a perfect square. This can be accomplished by solving the equation 0 = 16 - 4(2*z* + 10)(*z*^{2 }- 8), which is equivalent to *z*^{3} + 5*z*^{2 }- 8*z *- 42 = 0. By simple inspection we find a solution *z* = -3, therefore (*x*^{2 }- 3)^{2} = 4*x*^{2 }- 4*x* + 1; that is to say (*x*^{2} - 3)^{2} = (2*x *- 1)^{2}. Hence *x*^{2 }- 3 = 2*x *- 1 or (*x*^{2 }- 3) = -(2*x *- 1). The solutions of these quadratics are 1 ± √3 and -1 ± √5. These are, then, the four solutions of the given quartic.

In problem IX Cardano takes a slightly different path: the given equation is equivalent to *x*^{4} - 2*x*^{2} + 1 = 8*x*^{2 }- 4*x *- 7. Adding -2*bx*^{2} + *b*^{2} + 2*b* to both sides of the equality we get *x*^{4} - (2 + 2*b*)*x*^{2} + (*b* + 1)^{2} = 8*x*^{2 }- 2*bx*^{2 }- 4*x* + *b*^{2} + 2*b *-7, hence (*x*^{2 }- (*b* + 1))^{2} = (8 - 2*b*)*x*^{2 }- 4*x* + (*b*^{2} + 2*b *-7). The right hand side will be a perfect square if 4 = (8 - 2*b*)(*b*^{2} + 2*b *- 7), thus *b*^{3} + 30 = 2*b*^{2} + 15*b*. It is not hard to see that *b* = 2 is a solution to the cubic, therefore (*x*^{2 }- 3)^{2} = 4*x*^{2 }- 4*x* + 1 or, what is the same, (*x*^{2 }- 3)^{2} = (2*x *- 1)^{2}. Finally we get *x*^{2 }- 3 = 2*x *- 1 or *x*^{2 }- 3 = -(2*x *- 1), the same quadratic equations we found in the previous paragraph. Why did Cardano choose a more elaborate path? Probably he preferred to reach a resolvent cubic with a positive solution *b* = 2 instead of the negative solution *z* = -3 obtained when we followed the first path. Let us recall that Cardano does accept negative solutions to equations, but he calls them "fictitious".

At the end of Problem IX Cardano asks whether in choosing another solution of the resolvent cubic we would reach the same solutions of the original quartic equation. He leaves this task to the reader asserting: "If this operation delights you, you may go ahead and inquire into this for yourself." Following Cardano's advice, let us first find another solution of the resolvent cubic *z*^{3} + 5*z*^{2} - 8*z *- 42 = 0. Since *z* = -3 is a solution we can apply the method of long division and reach the equivalent equation (*z* + 3)(*z*^{2} + 2*z* - 14) = 0. The solutions of *z*^{2} + 2*z *- 14 = 0 are -1 ± √15, thus we have the three solutions of the cubic under discussion.

Let us choose *z* = √15 - 1. Replacing this value of *z* in (*x*^{2} + *z*)^{2} = (10 + 2*z*)*x*^{2 }- 4*x* + (*z*^{2} - 8), an equation that we introduced at the beginning of the discussion of Problem IX, we obtain (*x*^{2} + √15-1)^{2} = (8 + 2√15)*x*^{2 }- 4*x* + 8 - 2√15. *However, any quadratic polynomial px^{2}* +

We can then conclude that $$x^{2}+\sqrt{15}-1=\sqrt{8+2\sqrt{15}}\left(x-\frac{1}{4+\sqrt{15}}\right) \quad {\rm or} \quad

x^{2}+\sqrt{15}-1=-\sqrt{8+2\sqrt{15}}\left(x-\frac{1}{4+\sqrt{15}}\right).$$ The first quadratic equation has solutions √3 + 1 and √5 - 1, while the second quadratic equation has solutions – √3 + 1 and – √5 - 1. These are the four solutions of *x*^{4} + 4*x* + 8 = 10*x*^{2} found before, thus giving an affirmative answer to Cardano's question. Note that Cardano has wisely chosen problems XII and IX in such a way that the corresponding cubic has an integer solution.