The five problems about quartics that we have discussed so far (problems XII, IX, V, VI, and VIII) have no cubic terms; thus, Ferrari's method can be applied readily. In problem VII, namely *x*^{4 }+ 6*x*^{3 }= 64, Cardano realizes that a transformation is needed to eliminate the cubic term. Indeed, the seventh rule of chapter 7 suggests defining the transformation *y *= 4/*x*, which leads to *y*^{4 }= 6*y *+ 4. This is a quartic without cubic term, hence we can apply Ferrari's method to it; for each solution *y *= *b* we will get a root *x *= 4/*b* of the original equation. In problem VII Cardano does not look for the solutions, rather he is mainly interested in stressing the need to apply a transformation to eliminate the cubic term.

We may add that by inspection it is possible to conclude that *x *= 2 is a root of *x*^{4 }+ 6*x*^{3 }= 64, then by long division we get the equivalent equation (*x *- 2)(*x*^{3 }+ 8*x*^{2 }+ 16*x *+ 32) = 0. The problem has been reduced to the solution of a cubic.

Figure 5: Problem XI from chapter 39

Cardano solves problem XI, namely *x*^{4 }- 3*x*^{3 }= 64, in a rather novel way. Essentially, he does the following: A new variable *b* is introduced and the expression 2*bx*^{2 }+ (*b*^{2}/64)*x*^{4} is added to both sides of the aforementioned equation, leading to (*b*^{2}/64 + 1)*x*^{4 }- 3*x*^{3 }+ 2*bx*^{2 }= 64 + 2*bx*^{2 }+ (*b*^{2}/64)*x*^{4}. Consequently $$x^2left[left({b^2over 64} + 1
ight)x^2-3x+2b
ight] = left(8+{bover 8}x^2
ight)^2 (1)$$

The expression in brackets will be a perfect square provided (*b*^{2}/64+1)(2*b*) = (-3/2)^{2}, i.e. *b*^{3 }+ 64*b *= 72. The real (positive) solution of this cubic is $$b=
oot 3 of{sqrt{11,005{1over 27}}+36} -
oot 3of{sqrt{11,005{1over 27}}-36}.$$ Therefore $$left({b^2over 64} + 1
ight)x^2-3x+2b = left({b^2over 64}+1
ight)left(x - {3over 2(b^2/64 + 1)}
ight)^2 = left(xsqrt{{b^2over 64}+1} - {3over 2sqrt{b^2/64 + 1}}
ight)^2.$$ But 3/2=√(*b*^{2}/64+1)√(2*b)*, thus $$left({b^2over 64}+1
ight)x^2-3x+2b = left(xsqrt{{b^2over 64}+1}-sqrt{2b}
ight)^2.$$ Replacing this expression in (1) we obtain $$left(x^2sqrt{{b^2over 64}+1} - xsqrt{2b}
ight)^2 = left(8+{bover 8}x^2
ight)^2.$$ Therefore $$x^2sqrt{{b^2over 64}+1}-xsqrt{2b} = 8+{bover 8}x^2.$$ Cardano is not interested in going any further because his stated purpose is to show a methodology that does not require a transformation.

Another possible quadratic is $$x^2sqrt{{b^2over 64}+1} - xsqrt{2b} = -left(8+{bover 8}x^2
ight),$$ which does not appear in *Ars Magna*. Of course, we can provide the solutions of both quadratics using the decimal approximation 1.10398 for *b;* the first quadratic has the solutions 4 and -2.29495 while the second has complex solutions. It should come as no surprise that 4 is a solution because by simple inspection we could have noticed, right at the beginning, that *x *= 4 is indeed a solution of *x*^{4 }- 3*x*^{3 }= 64. Then long division leads to (*x *- 4)(*x*^{3 }+ *x*^{2 }+ 4*x *+ 16) = *x*^{4 }- 3*x*^{3 }- 64 and we can ascertain that the only real solution of *x*^{3 }+ *x*^{2 }+ 4*x *+ 16 = 0 is approximately -2.29495.

Cardano does mention that Problem XI can be solved through the same transformation used in Problem VII, though he does not pursue this path. Indeed, defining *y *= 4/*x* we get the quartic equation *y*^{4 }= -3*y *+ 4 and we realize that for any *z* we have (*y*^{2 }+ *z*)^{2 }= 2*zy*^{2 }- 3*y *+ (*z*^{2 }+ 4). The expression to the right of the equal sign will be a perfect square if 0 = 9 - 8*z*(*z*^{2 }+ 4), that is to say *z*^{3 }= - 4*z *+ 9/8. The usual procedure for cubics leads to the real solution $$z=
oot 3of{{9over 16}+{1over 2}sqrt{{81over 64}+{256over 27}}}-
oot 3of {-{9over 16}+{1over 2}sqrt{{81over 64}+{256over 27}}} approx 0.275994.$$ Then $$(y^2+0.275994)^2 = 2cdot 0.275994y^2-3y + (0.275994^2+4) = 0.551988left(y-{3over 2cdot 0.551988}
ight)^2.$$ Therefore $$y^2+0.2759994 = pmsqrt{0.551988}left(y-{3over 1.10398}
ight).$$ The solutions of the first quadratic are complex, but the solutions of the second quadratic are approximately 0.999996 and -1.74295. Then *x *= 4/0.999996 = 4.00002 and *x *= 4/(-1.74295) = -2.29496. The first solution is, as expected, practically 4.

The last problem of chapter 39 is Problem XIII; after that there are no more references to quartics in* Ars Magna*. Problem XIII states: " Find a number the fourth power of which plus twice its cube is one more than the number." Thus, it is necessary to solve the equation *x*^{4 }+ 2*x*^{3 }= *x *+ 1. Evidently, a transformation of the type *y *= *c*/*x* will not eliminate the cubic term. Cardano chooses an ad-hoc method, which can be interpreted as follows: the given equation is equivalent to *x*^{4 }+ 2*x*^{3 }+ *x*^{2 }= *x*^{2 }+ *x *+ 1, thus (*x*^{2 }+ *x*)^{2 }= 1 + *x *+ *x*^{2}. Letting *y *= *x*^{2 }+ *x* we get *y*^{2 }= *y *+ 1. One solution of this equation is *y*=1/2+√5/2. In turn, the solutions of *x*^{2 }+ *x *= 1/2 + √5/2 are -1/2 ± √(3/4 + √5/2) (only the positive solution is mentioned in *Ars Magna*). It is worth mentioning that the other solution of *y*^{2 }= *y *+ 1, namely *y*=1/2 - √5/2, leads to the two complex solutions -1/2 ± *i*√(√5/2 - 3/4) of the original problem.

As an alternative path, not pursued by Cardano, we can use the transformation *z *= *x *+ 2/4, i.e. *x *= *z *- 1/2, which eliminates the cubic term of the equation *x*^{4 }+ 2*x*^{3 }= *x *+ 1 and leads to the biquadratic equation *z*^{4 }- (3/2)*z*^{2 }- 11/16 = 0, whose four solutions are $$pmsqrt{{3over 4}+{sqrt{5}over 2}}, quad pm isqrt{{sqrt{5}over 2}-{3over 4}}.$$ Then the four solutions of the original equation are $$-{1over 2}pm sqrt{{3over 4}+{sqrt{5}over 2}}, quad -{1over 2}pm isqrt{{sqrt{5}over 2}-{3over 4}}.$$

Having exhausted the contributions of Cardano and Ferrari to the problem of quartics in chapter 39, in the next section we will analyze an approach to biquadratics that was developed almost 200 years after the publication of the first edition of *Ars Magna* at a time when symbolic algebra had already reached its present form.