Elements of Algebra

CHAP. VII. OF ALGEBRA. 231 $x=-\frac{1}{2}+\sqrt{(\frac{1}{4}+182)}=-\frac{1}{2}+\sqrt{(\frac{729}{4})}=-\frac{1}{2}+\frac{27}{2}=13$; from which we conclude, that the triangular root required is 13; for the triangle of 13, or $\frac{x^2+x}{2}$ is 91. 659. But, in general, let $a$ be the given triangular number, and let its root be required. Here, if we make it $=x$, we have $\frac{x^2+x}{2}=a$, or $x^2+x=2a$; therefore, $x^2=-x+2a$, and, by the rule for solving Quadratic Equations [Art. 641.] $x=-\frac{1}{2}+\sqrt{(\frac{1}{4}+2a)}$, or $x=\frac{-1+\sqrt{(8a+1)}}{2}$. This result gives the following rule: To find a triangular root, Multiply the given triangular number by 8, add 1 to the product, extract the root of the sum, subtract 1 from that root, and lastly, divide the remainder by 2. 660. So that all triangular numbers have this property; namely, if we multiply them by 8, and add unity to the product, the sum is always a square; of which the following small Table furnishes some examples: Triangles $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots$ 8 times $+1=9, 25, 49, 81, 121, 169, 225, 289, 361, 441, \ldots$ If the given number $a$ does not answer this condition, we conclude, that it is not a real triangular number, or that no rational root of it can be assigned. 661. According to this rule, let the triangular root of 210 be required; we shall have $a=210$, and $8a+1=1681$, the square root of which is 41; whence we see, that the number 210 is really triangular, and that its root is $\frac{41-1}{2}=20$. But if 4 were given as the triangular number, and its root were required, we should find it $=\frac{\sqrt{33}}{2}$, and consequently irrational. However, the triangle of this root, $\frac{\sqrt{33}}{2}-\frac{1}{2}$, may be found in the following manner: Since $x=\frac{\sqrt{33}-1}{2}$, we have $x^2=\frac{17-\sqrt{33}}{2}$, and adding