Elements of Algebra

CHAP. IV. OF ALGEBRA. 333 $3 + \frac{my}{n}$, and we shall thus obtain the equation $9 + 14y + 7y^2 = 9 + \frac{6my}{n} + \frac{m^2y^2}{n^2}$, in which we may expunge 9 from both sides, and divide by $y$: which being done, we shall have $14n^2 + 7n^2y = 6mn + m^2y$; whence $y = \frac{6mn-14n^2}{7n^2-m^2}$; and, consequently, $x = \frac{6mn-7n^2-m^2}{7n^2-m^2}$*, in which we may substitute any values we please for $m$ and $n$. If we make $m = 1$, and $n = 1$, we have $x = -\frac{2}{6}$; or, since the second power of $x$ stands alone, $x = +\frac{1}{3}$; wherefore $2 + 7x^2 = \frac{25}{9}$. If $m = 3$, and $n = 1$, we have $x = -1$, or $x = +1$. But if $m = 3$, and $n = -1$, we have $x = 17$; which gives $2 + 7x^2 = 2025$, the square of 45. If $m = 8$, and $n = 3$, we shall then have, in the same manner, $x = -17$, or $x = +17$. But, by making $m = 8$, and $n = -3$, we find $x = 271$; so that $2 + 7x^2 = 514089 = 717^2$. 60. Let us now examine the formula $5x^2 + 3x + 7$, which becomes a square by the supposition of $x = -1$. Here, if we make $x = y - 1$, our formula will be changed into this: $5y^2 - 7y + 9$, the square root of which we will suppose to be $3 - \frac{my}{n}$; by which means we have $5y^2 - 7y + 9 = 9 - \frac{6my}{n} + \frac{m^2y^2}{n^2}$, or $5n^2y - 7n^2 = -6mn + m^2y$; whence, $y = \frac{7n^2-6mn}{5n^2-m^2}$; and, lastly, $x = \frac{2n^2-6mn+m^2}{5n^2-m^2}$. * Because $x$ was made $= 1 + y$; and 1 is here added to the fractional expression, $\frac{6mn-14n^2}{7n^2-m^2}$.