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 *x*^{4 }+ *d*_{1}*x*^{3 }+ *d*_{2}*x*^{2 }+ *d*_{3}*x *+ *d*_{4 }= 0, where the coefficients are real numbers. Using the transformation *x*=*y *- *d*_{1}/4 we can convert it into an equation of the type *y*^{4 }+ *ay*^{2 }+ *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 *y*^{4 }= -*ay*^{2 }- *by *- *c*. So (*y*^{2 }+ *z*)^{2 }= 2*y*^{2}*z *+ *z*^{2 }- *ay*^{2 }- *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 (2*z *- *a*)*y*^{2 }- *by *+ (*z*^{2 }- *c*) will be a perfect square provided its discriminant is zero, i.e. *b*^{2 }- 4(2*z *- *a*)(*z*^{2 }- *c*) = 0 or, what is the same, 8*z*^{3 }- 4*az*^{2 }- 8*cz *+ (4*ac *- *b*^{2}) = 0. This is a cubic equation, customarily called a "resolvent cubic". Let *z*_{1} be a real solution of it. Evidently 2*z*_{1 }- *a* ¹ 0 because otherwise the resolvent cubic implies *b*^{2 }= 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 *y*^{4 }+ *ay*^{2 }+ *by *+ *c *= 0, which in turn lead to the four solutions of the original quartic through the use of the transformation *x *= *y *- *d*_{1}/4. We have to keep in mind that if 2*z*_{1 }- *a* < 0 we will have to deal with a quadratic that happens to have complex coefficients.

An alternative way to solve *y*^{4 }+ *ay*^{2 }+ *by *+ *c *= 0, through Ferrari's technique, starts by noticing that (*y*^{2 }+ *a*/2)^{2 }= -*by *-*c *+ *a*^{2}/4 (recall Cardano's solution of problem VIII, discussed in detail in section 6). For any *z* we will have (*y*^{2 }+ *a*/2 + *z*)^{2 }= -*by *- *c *+ *a*^{2}/4 + *z*^{2 }+ 2*z*(*y*^{2 }+ *a*/2), that is to say (*y*^{2 }+ *a*/2 + *z*)^{2 }= 2*zy*^{2 }- *by *+ (*z*^{2}- *c *+ *a*^{2}/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 *b*^{2 }- 8*z*(*z*^{2 }- *c *+ *a*^{2}/4 + *az*) = 0. In other words, we need to solve the equation 8*z*^{3 }+ 8*az*^{2 }- 8*cz *+ 2*a*^{2}*z *- *b*^{2 }= 0, a resolvent cubic different from the one we found before.

Let *z*_{1} be a real solution of it, which has to be different from zero because otherwise *b *= 0; therefore (*y*^{2 }+ *a*/2 + *z*_{1})^{2 }= 2*z*_{1}(*y *- *b*/4*z*_{1})^{2}. The four solutions of *y*^{4 }+ *ay*^{2 }+ *by *+ *c *= 0 will stem from the quadratics (*y*^{2 }+ *a*/2 + *z*_{1}) = √(2*z*_{1})(*y *- *b*/4*z*_{1}), (*y*^{2 }+ *a*/2 + *z*_{1}) = -√(2*z*_{1)}(*y *- *b*/4*z*_{1}). 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 *x*^{4 }+ 12 = 6*x*^{2}, whose complex solutions were found at the end of the previous section. We have (*x*^{2}+*z*)^{2 }= 6*x*^{2 }- 12 + *z*^{2 }+ 2*zx*^{2} 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(2*z *+ 6)(*z*^{2 }- 12)=0. We can choose the solution √12. Replacing this value in (3) we get (*x*^{2 }+ 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.