Elements of Algebra

232 ELEMENTS SECT. IV. $x = \frac{\sqrt{33}-1}{2}$ to it, the sum is $x^2 + x = \frac{16}{2} = 8$. Consequently, the triangle, or the triangular number, $\frac{x^2+x}{2} = 4$. 662. The quadrangular numbers being the same as squares, they occasion no difficulty. For, supposing the given quadrangular number to be $a$, and its required root $x$, we shall have $x^2 = a$, and consequently, $x = \sqrt{a}$; so that the square root and the quadrangular root are the same thing. 663. Let us now proceed to pentagonal numbers. Let 22 be a number of this kind, and $x$ its root; then, by the third formula, we shall have $\frac{3x^2-x}{2} = 22$, or $3x^2 - x = 44$, or $x^2 = \frac{1}{3}x + \frac{44}{3}$; from which we obtain, $x = \frac{1}{6} + \sqrt{(\frac{1}{36} + \frac{44}{3})}$, or $x = \frac{1+\sqrt{(529)}}{6} = \frac{1}{6} + \frac{23}{6} = 4$; and consequently 4 is the pentagonal root of the number 22. 664. Let the following question be now proposed; the pentagon $a$ being given, to find its root. Let this root be $x$, and we have the equation $\frac{3x^2-x}{2} = a$, or $3x^2 - x = 2a$, or $x^2 = \frac{1}{3}x + \frac{2a}{3}$; by means of which we find $x = \frac{1}{6} + \sqrt{(\frac{1}{36} + \frac{2a}{3})}$, that is, $x = \frac{1+\sqrt{(24a+1)}}{6}$. Therefore, when $a$ is a real pentagon, $24a+1$ must be a square. Let 330, for example, be the given pentagon, the root will be $x = \frac{1+\sqrt{(7921)}}{6} = \frac{1+89}{6} = 15$. 665. Again, let $a$ be a given hexagonal number, the root of which is required. If we suppose it $x$, we shall have $2x^2 - x = a$, or $x^2 = \frac{1}{2}x + \frac{1}{2}a$; and this gives $x = \frac{1}{4} + \sqrt{(\frac{1}{16} + \frac{1}{2}a)} = \frac{1+\sqrt{(8a+1)}}{4}$. So that, in order that $a$ may be really a hexagon, $8a+1$ must become a square; whence we see, that all hexagonal numbers are contained in triangular numbers; but it is not the same with the roots.