Elements of Algebra

CHAP. IX. OF ALGEBRA. 377 $q=4$; whence $y=0$, and $x=1$; which, as before, leads to nothing new. 144. Again, let the formula be $1+8x^2+x^4$; in which $a=1$, $c=8$, and $e=1$. Here a slight consideration is sufficient to point out the satisfactory case, namely, $x=2$; for, by supposing $h=2$, we find $k=7$; so that making $x=2+y$, and representing the root by $7+py+qy^2$, we shall have $p=\frac{32}{7}$, and $q=\frac{272}{343}$; whence $y=-\frac{5880}{2911}$, and $x=-\frac{58}{2911}$; and we may omit the sign minus in these values. But we may observe, farther, in this example, that, since the last term is already a square, and must therefore remain a square also in the new formula, we may here apply the method which has been already taught for cases of the third class. Therefore, as before, let $x=2+y$, and we shall have $1$ $32+32y+8y^2$ $16+32y+24y^2+8y^3+y^4$ $49+64y+32y^2+8y^3+y^4$, an expression which we may now transform into a square in several ways. For, in the first place, we may suppose the root to be $7+py+y^2$; and, consequently, the formula equal to the square $49+14py+p^2y^2+14y^2+2py^3+y^4$; but then, after destroying $8y^3$, and $2py^3$, by supposing $2p=8$, or $p=4$, dividing the other terms by $y$, and deriving from the equation, $64+32y=14p+14y+p^2y=56+30y$, the value of $y=-4$, and of $x=-2$, or $x=+2$, we come only to the case that is already known. Farther, if we seek to determine such a value for $p$, that the second terms may vanish, we shall have $14p=64$, and $p=\frac{32}{7}$; and the other terms, when divided by $y^2$, form the equation $14+p^2+2py=32+8y$, or $\frac{1710}{49}+\frac{64}{7}y=32+8y$, whence we find $y=-\frac{71}{28}$; and, consequently, $x=-\frac{15}{28}$, or $x=+\frac{15}{28}$; and this value transforms our formula into a square, whose root is $\frac{1441}{784}$. Farther, as $-y^2$ is no less the root of the last term than $+y^2$, we may suppose the root of the formula to be $7+py-y^2$, or the formula itself equal to $49+14py+p^2y^2-14y^2-2py^3+y^4$. And here we shall destroy the last terms but one, by making $-2p=8$, or $p=-4$; then, dividing the other terms by $y$, we shall have