Elements of Algebra

370 ELEMENTS PART II. $ay^4 + by^3 + cy^2 + dy + g^4$, which is quite similar to the former, only inverted. But it is not necessary to go through this process; we have only to suppose the root to be $gx^2 + px + q$, or, inversely, $q + px + gx^2$, and we shall thus have $a + bx + cx^2 + dx^3 + g^2x^4 = q^2 + 2pqx + 2gqx^2 + p^2x^2 + 2gpx^3 + g^2x^4$. Now, the fifth and sixth terms destroying each other, we shall first determine $p$ so, that the fourth terms may also destroy each other; which happens when $d = 2gp$, or $p = \frac{d}{2g}$; we shall then likewise determine $q$, in order to remove the third terms, making for this purpose $c = 2gq + p^2$, or $q = \frac{c-p^2}{2g}$; which done, the first two terms will furnish the equation $a + bx = q^2 + 2pqx$; whence we obtain $x = \frac{a-q^2}{2pq-b}$, or $x = \frac{q^2-a}{b-2pq}$. 133. Here, again, we find the same imperfection that was before remarked, in the case where the second and fourth terms are wanting; that is to say, $b = 0$, and $d = 0$; because we then find $p = 0$, and $q = \frac{c}{2g}$; therefore $x = \frac{a-q^2}{0}$: now, this value being infinite, leads no farther than the value, $x = 0$, in the first case; whence it follows, that this method cannot be at all employed with respect to expressions of the form $a + cx^2 + g^2x^4$. 134. 3d. Resolution of the formula $\sqrt{(f^2 + bx + cx^2 + dx^3 + g^2x^4)}$. It is evident that we may employ for this formula both the methods that have been made use of; for, in the first place, since the first term is a square, we may assume $f + px + qx^2$ for the root, and make the first three terms vanish; then, as the last term is likewise a square, we may also make the root $q + px + gx^2$, and remove the last three terms; by which means we shall find even two values of $x$. But this formula may be resolved also by two other methods, which are peculiarly adapted to it. In the first, we suppose the root to be $f + px + gx^2$, and