Elements of Algebra

CHAP. IV. OF ALGEBRA. 327 sible numbers; but we shall generally find a fraction for $x$, though sometimes we may obtain an integer number. For example, if $m=3$, and $n=2$, we find $x=8$, the triangular number of which, or 36, is also a square. We may also make $m=7$, and $n=5$; in this case, $x=50$, the triangle of which, 1225, is at the same time the triangle of $+49$, and the square of 35. We should have obtained the same result by making $n=7$ and $m=10$; for, in that case, we should also have found $x=49$. In the same manner, if $m=17$ and $n=12$, we obtain $x=288$, its triangular number is $\frac{x(x+1)}{2} = \frac{288 \times 289}{2} = 144 \times 289$, which is a square, whose root is $12 \times 17 = 204$. 50. We may remark, with regard to this last case, that we have been able to transform the formula $bx+cx^2$ into a square from its having a known factor, $x$. This observation leads to other cases, in which the formula $a+bx+cx^2$ may likewise become a square, even when neither $a$ nor $c$ is a square. These cases occur when $a+bx+cx^2$ may be resolved into two factors; and this happens when $b^2-4ac$ is a square: to prove which, we may remark, that the factors depend always on the roots of an equation; and that, therefore, we must suppose $a+bx+cx^2=0$. This being laid down, we have $cx^2 = -bx-a$, or $x^2 = -\frac{bx}{c} - \frac{a}{c}$, whence, by compleating the square, &c., we find $x = -\frac{b}{2c} \pm \sqrt{(\frac{b^2}{4c^2} - \frac{a}{c})}$, or $x = -\frac{b}{2c} \pm \frac{\sqrt{(b^2-4ac)}}{2c}$, and, it is evident, that if $b^2-4ac$ be a square, this quantity becomes rational. Therefore let $b^2-4ac=d^2$; then the roots will be $\frac{-b \pm d}{2c}$, that is to say, $x = \frac{-b+d}{2c}$; and, consequently, the divisors of the formula $a+bx+cx^2$ are $x+\frac{b-d}{2c}$, and $x+\frac{b+d}{2c}$. If we multiply these factors together, we shall be brought to the same formula again, except that it is divided by $c$; for the product is $x^2 + \frac{bx}{c} + \frac{b^2}{4c^2} - \frac{d^2}{4c^2}$; and since $d^2 = b^2-4ac$, we have