Elements of Algebra

CHAP. IX. OF ALGEBRA. 371 $p$ is determined such, that the second terms destroy each other; that is to say, $f^2 + bx + cx^2 + dx^3 + g^2x^4 = f^2 + 2fpx + 2fgx^2 + p^2x^2 + 2gpx^3 + g^2x^4$. Then, making $b = 2fp$, or $p = \frac{b}{2f}$; and since by these means both the second terms, and the first and last, are destroyed, we may divide the others by $x^2$, and shall have the equation $c + dx = 2fg + p^2 + 2gpx$, from which we obtain $x = \frac{c-2fg-p^2}{2gp-d}$, or $x = \frac{p^2+2fg-c}{d-2gp}$. Here, it ought to be particularly observed, that as $g$ is found in the formula only in the second power, the root of this square, or $g$, may be taken negatively as well as positively; and, for this reason, we may obtain also another value of $x$; namely, $x = \frac{c+2fg-p^2}{-2gp-d}$, or $x = \frac{p^2-2fg-c}{2gp+d}$. 135. There is, as we observed, another method of resolving this formula; which consists in first supposing the root, as before, to be $f+px+gx^2$, and then determining $p$ in such a manner, that the fourth terms may destroy each other; which is done by supposing in the fundamental equation, $d=2gp$, or $p = \frac{d}{2g}$; for, since the first and the last terms disappear likewise, we may divide the other by $x$, and there will result the equation $b + cx = 2fp + 2fgx + p^2x$, which gives $x = \frac{b-2fp}{2fg+p^2-c}$. We may farther remark, that as the square $f^2$ is found alone in the formula, we may suppose its root to be $-f$, from which we shall have $x = \frac{b+2fp}{p^2-2fg-c}$. So that this method also furnishes two new values of $x$; and, consequently, the methods we have employed give, in all, six new values. 136. But here again the inconvenient circumstance occurs, that, when the second and the fourth terms are wanting, or when $b=0$, and $d=0$, we cannot find any value of $x$ which answers our purpose; so that we are unable to resolve the formula $f^2 + cx^2 + gx^4$. For, if $b=0$, and BB2