ON ARITHMETIC BOOK X.

8

But when the square and the number are equal to things Modern form: $x^2 + n = ax$, and the case is possible, then there are two true solutions original: "duæ ſolutiones ueræ," meaning two positive roots, as by saying the square plus 12 is equal to 7 positions $x^2 + 12 = 7x$, the position can be 4 or even 3. For it is verified in both, except when the number is equal to the square of half the number of roots i.e., when $n = (a/2)^2$, resulting in a single repeated root, for then there is only one equation, namely half of the number of those roots. In this chapter, however, there can never be a feigned solution feigned solution original: "solutio ficta," Cardano's term for a negative root, nor an equation through "minus." But where there is a solution through a true number, it is double; where it lacks a true solution, it cannot be solved any further by a feigned equation.

8

If indeed an equation is sought in the chapters of the cube, squares, and the number, then if the cube is equal to the squares and the number $x^3 = ax^2 + n$, then there is only one true solution, just as if I say, the cube is equal to three squares plus 16 $x^3 = 3x^2 + 16$, the thing is worth 4, and no other can be found.

NOTE

NOTE WELL. In all chapters in which there is only one solution, the equation is easier to find and more elegant, such as in the chapter of the cube and things equal to a number $x^3 + ax = n$, and the cube equal to the square and the number $x^3 = ax^2 + n$, and in the chapter of the cube equal to things and a number $x^3 = ax + n$, where that product of $\frac{2}{3}$ of the number of things into the square root of a third part of the things is less than the number. I say the same where the cube with the number is equal to things $x^3 + n = ax$, and only a feigned equation can be had; the others, however, in which the estimation estimation original: "æstimatio," the value of the unknown or the root of the thing is manifold, are more difficult and confused.

If therefore the cube and the square are equal to a number $x^3 + ax^2 = n$, then the estimation of the thing is only one through "plus" a positive root, where from $\frac{1}{3}$ of the number of the square into the square of two-thirds of the same number, there is produced less than the number of the equation. And this same estimation is the feigned one of the corresponding chapter of the cube and the number equal to squares under the same quantity. Example: The cube and three squares are equal to 20 $x^3 + 3x^2 = 20$; then, because from 1 (the third part of the number of squares) into 4 (the square of two-thirds) it becomes less than 20, I say that there is only one equation, and the thing is worth 2; and this is the estimation through "minus" the negative root of the cube plus 20 being equal to three squares $x^3 + 20 = 3x^2$. But where such a number can be produced from that multiplication, there will be one true estimation and two feigned ones, and the true will correspond to the feigned of the other chapter, and again the feigned to the true ones. Example: if I say, the cube and 11 squares are equal to 72 $x^3 + 11x^2 = 72$, the thing is the square root of 40 minus 4 $\sqrt{40} - 4$ for the true estimation, but for the feigned it is 3 minus $-3$ or the square root of 40 plus 4 minus $-(\sqrt{40} + 4)$. And if the cube with 72 is equal to 11 squares $x^3 + 72 = 11x^2$, the true estimations are 3 or the square root of 40 plus 4, and the feigned is the square root of 40 minus 4 minus $-(\sqrt{40} - 4)$. Therefore, in seeking the feigned solution, we always seek the true one and the corresponding one of the other chapter.

Note

It is known from this, moreover, that certain chapters have two estimations, some one, and when it has three, in one part of the ca...