Problem VI asks to "find a number which is equal to its square root plus twice its cube root." Denoting the positive number by *x*^{6}, we have to solve the equation *x*^{6}= *x*^{3 }+ 2*x*^{2}. Dividing by *x*^{2} we get the equivalent equation *x*^{4}= *x *+ 2, a quartic indeed. Cardano subtracts 1 from each side and reaches *x*^{4 }- 1= *x *+ 1, thus (*x*^{4 }- 1)/(*x *+ 1)= 1. Consequently *x*^{3 }- *x*^{2}+ *x *- 1= 1, that is to say *x*^{3 }+ *x *= *x*^{2 }+ 2. He ascertains that $$x=\root 3\of{\sqrt{83\over 108}+{47\over 54}}-\root 3\of{\sqrt{83\over 108}-{47\over 54}}+{1\over 3}$$is a solution, without providing the details. Let us check whether this is a solution, using for that purpose the Cardano-Tartaglia method.

Defining *y *= *x*-1/3 we reach the depressed cubic *y*^{3 }= -(2/3)*y*+ 47/27. Keeping in mind that (*u *- *v*)^{3 }= -3*uv*(*u *- *v*) + (*u*^{3 }- *v*^{3}), a solution of the form *y *= *u *- *v* will be found provided that the algebraic system -3*uv*= -2/3, *u*^{3 }- *v*^{3}= 47/27 has a solution. Indeed *u*^{3}- (2/9*u*)^{3}= 47/27, that is to say *u*^{6 }- (47/27)*u*^{3 }- 8/729 = 0. One solution of this equation is $$u^3 = {47\over 54} + {1\over 2}\sqrt{\left({47\over 27}\right)^2 + {32\over 729}},$$ thus $$u = \root 3\of {{47\over 54}+\sqrt{83\over 108}}.$$ Replacing this value of *u* in *u*^{3 }- *v*^{3}= 47/27 we get $$v=\root 3\of{-{47\over 54}+\sqrt{83\over 108}}.$$ Then $$y=u-v=\root 3\of{{47\over 54}+\sqrt{83\over 108}} - \root 3\of{-{47\over 54}+\sqrt{83\over 108}},$$ which in turn leads to $$x=\root 3\of{\sqrt{83\over 108} + {47\over 54}}-\root 3\of{\sqrt{83\over 108}-{47\over 54}}+{1\over 3}.$$The other two solutions of the cubic are complex numbers, no wonder that Cardano ignores them.

A natural question, addressed in *Ars Magna*, is what happens if we follow Ferrari's procedure: We start with (*x*^{2})^{2 }= *x *+ 2 and introduce a new variable *b*. We observe that (*x*^{2 }+ *b*)^{2 }= 2*bx*^{2 }+ *x *+ *b*^{2 }+ 2. In order to have a perfect square we need to impose the condition 1/4 = 2*b*(*b*^{2 }+ 2), equivalent to demanding that the discriminant of 2*bx*^{2 }+ *x *+ *b*^{2 }+ 2 is zero. Thus *b*^{3 }+ 2*b*= 1/8. Cardano ascertains that $$b=\root 3\of{{1\over 16}+\sqrt{2,075\over 6,912}}-\root 3\of{-{1\over 16}+\sqrt{2,075\over 6,912}}$$is a solution. We would have to replace this value of *b*, call it *b*_{1}, in (*x*^{2 }+ *b*)^{2}= 2*bx*^{2 }+ *x *+ *b*^{2 }+ 2 knowing beforehand that the expression to the right is also a perfect square. In other words, it is necessary to solve (*x*^{2 }+ *b*_{1})^{2 }= 2*b*_{1}(*x*+ 1/4*b*_{1})^{2} or, equivalently, the pair of quadratic equations $$x^2+b_1=\sqrt{2b_1}\left(x+{1\over 4b_1}\right), \qquad x^2+b_1=-\sqrt{2b_1}\left(x+{1\over 4b_1}\right).$$The solutions of the first quadratic are $$x=\sqrt{b_1\over 2} \pm\sqrt{{1\over 2\sqrt{2b_1}}-{b_1\over 2}}.$$

Using the decimal approximation 0.062379 to *b*_{1} we get *x *= 1.35321 and *x *= -0.999998. The first is a decimal approximation to the solution found before, while the other root is a decimal approximation to -1. But we are not interested in the latter because we are seeking only positive solutions to the problem (from the very beginning we ruled out the trivial solutions of the equation *x*^{6 }= *x*^{3 }+ 2*x*^{2}, namely 0 and -1). It should be noted that the quadratic equation \( x^2 +b_1 = -\sqrt{2b_1} \left( x+\frac{1}{4b_1} \right) \) has complex roots.

Problem VIII asks to "divide 6 into three proportional quantities the sum of the squares of the first and second of which is 4." Let *x*, *y*, *z* denote the quantities. Since *x*^{2 }+ *y*^{2}= 4 we will have \( y = \sqrt{4 - x^2} \). But *x *+ *y *+ *z *= 6, consequently \( z = 6 - x - \sqrt{4 - x^2} \). Next Cardano ascertains that \( \left(6 - x - \sqrt{4 - x^2}\right) x = {4 - x^2} \), which is correct since *x*, *y*, *z* are proportional quantities. Then \( 6x - x^2 - x\sqrt{4 - x^2} = {4 - x^2} \), that is to say \( 6x - 4 = x\sqrt{4 - x^2} \). Squaring both sides we get *x*^{4 }+ 32*x*^{2 }+ 16 = 48*x*, which is equivalent to (*x*^{2 }+ 16)^{2 }= 48*x *+ 240.

For any *b* we have (*x*^{2 }+ 16 + *b*)^{2 }= 2*bx*^{2 }+ 48*x *+ *b*^{2 }+ 32*b *+ 240. We need to have the discriminant equal to zero in order to make sure that the right hand side is a perfect square, thus 0= 48^{2 }- 8*b*(*b*^{2 }+ 32*b *+ 240). This expression is equivalent to *b*^{3 }+ 32*b*^{2 }+ 240*b *= 288. Then Cardano uses the transformation *c*= *b *+ 32/3 and reaches the equation *c*^{3 }= (101 1/3)*c *+ 420 20/27, with a real solution *c*. Therefore *b *= *c *- 32/3 is a real solution of the resolvent cubic. Actually, Cardano does not introduce a new variable explicitly. Finally we have (*x*^{2 }+ 16+ *b*)^{2 }= 2*b*(*x *+ 48/4*b*)^{2}, which in turn leads to (*x*^{2 }+ 16+ *b*)= ±√(2*b)*(*x*+ 12/*b*). Cardano only considers the equation with positive sign and writes it as *x*^{2 }+ 16 + *b *= √(2*b)**x*+ √(*b*^{2 }+ 32*b *+ 240), which is in agreement with *x*^{2 }+ 16 + *b *= √(2*b*)*x*+ √(2*b)(*12/*b*) because √(2*b)(*12/*b*) = √(288/*b*)= √[(*b*^{3 }+ 32*b*^{2 }+ 240*b*)/*b*] = √(*b*^{2 }+ 32*b *+ 240). He leaves the task of finding *b*, and solving the above-mentioned quadratic equation, to the readers of *Ars Magna*.

An alternative path is to start with *x*^{4 }= -32*x*^{2 }+ 48*x *- 16 and introduce a new variable *z* so that the expression on the right of (*x*^{2}+ *z*)^{2 }= (2*z *- 32)*x*^{2 }+ 48*x *+ *z*^{2 }- 16 is a perfect square. This can be accomplished by solving 0 = 48^{2 }- 4(2*z *- 32)(*z*^{2 }-16), i.e. *z*^{3 }- 16*z*^{2 }- 16*z *- 32= 0. The usual approach to cubics leads to \( z \approx 17.0486 \), therefore $$(x^2+17.0486)^2 = 2.0972x^2+48x+274.655 = 2.0972\left(x+{48\over 2\cdot 2.0972}\right)^2, \quad {\rm or}\quad (x^2+17.0486)^2 = \left(\sqrt{2.0972}\left(x+{48\over 4.1944}\right)\right)^2.$$ Hence $$x^2+17.0486 = \sqrt{2.0972}\left(x+{48\over 4.1944}\right) \quad {\rm or} \quad x^2+ 17.0486 = -\sqrt{2.0972}\left(x+{48\over 4.1944}\right).$$ The first quadratic has the real solutions 0.943916 and 0.504855 while the second quadratic has complex solutions. Obviously, both real solutions are approximations.