Elements of Algebra

CHAP. IX. OF ALGEBRA. 369 130. If, therefore, the root be $f + px + qx^2$, and for that reason $f^2 + bx + cx^2 + dx^3 + ex^4 = f^2 + 2fpx + p^2x^2 + 2fqx^2 + 2pqx^3 + q^2x^4$, the first terms disappear of themselves; with regard to the second, we shall remove them by making $b = 2fp$, or $p = \frac{b}{2f}$; and, for the third, we must make $c = 2fq + p^2$, or $q = \frac{c-p^2}{2f}$. This being done, the other terms will be divisible by $x^3$, and will give the equation $d + ex = 2pq + q^2x$, from which we find $x = \frac{d-2pq}{q^2-e}$, or $x = \frac{2pq-d}{e-q^2}$. 131. Now, it is easy to see that this method leads to nothing, when the second and third terms are wanting in our formula; that is to say, when $b=0$, and $c=0$; for then $p=0$, and $q=0$; consequently, $x = -\frac{d}{e}$, from which we can commonly draw no conclusion, because this case evidently gives $dx^3 + ex^4 = 0$; and, therefore, our formula becomes equal to the square $f^2$. But it is chiefly with respect to such formulae as $f^2 + ex^4$, that this method is of no advantage, since in this case we have $d=0$, which gives $x=0$, and this leads no farther. It is the same, when $b=0$, and $d=0$; that is to say, the second and fourth terms are wanting, in which case the formula is $f^2 + cx^2 + ex^4$; for, then $p=0$, and $q = \frac{c}{2f}$, whence $x=0$, as we may immediately perceive, from which no further advantage can result. 132. 2d. Resolution of the formula $\sqrt{a + bx + cx^2 + dx^3 + g^2x^4}$. We might reduce this formula to the preceding case, by supposing $x = \frac{1}{y}$; for, as the formula $a + \frac{b}{y} + \frac{c}{y^2} + \frac{d}{y^3} + \frac{g^2}{y^4}$ must then be a square, and remain a square if multiplied by the square $y^4$, we have only to perform this multiplication, order to obtain the formula BB