Elements of Algebra

374 ELEMENTS PART II. If, therefore, we substitute this value of $x$ in the formula $a + ex^4$, it becomes a square; and its root, which we have supposed to be $k + py + qy^2$, will have this form, $k + \frac{8k(k^2-a) \times (2a-k^2)}{3k^4-4a^2} + \frac{16k(k^2-a) \times (k^2+2a) \times (2a-k^2)^2}{(3k^4-4a^2)^2}$; because, as we have seen, $p = \frac{2eh^3}{k}$, $q = \frac{eh(k^2+2a)}{k^3}$, $y = \frac{4hk^2(2a-k^2)}{3k^4-4a^2}$, and $eh^4 = k^2-a^*$. 139. Let us continue the investigation of the formula $a + ex^4$; and, since the case $a + eh^4 = k^2$ is known, let us consider it as furnishing two different cases; because $x = +h$, and $x = -h$; for which reason we may transform our formula into another of the third class, in which the first term and the last are squares. This transformation is made by an artifice, which is often of great utility, and which consists in making $x = \frac{h(1+y)}{1-y}$: by which means the formula becomes $\frac{a(1-y)^4+eh^4(1+y)^4}{(1-y)^4}$, or rather $\frac{k^2+4(k^2-2a)y+6k^2y^2+4(k^2-2a)y^3+k^2y^4}{(1-y)^4}$. Now, let us suppose the root of this formula, according to the third case, to be $\frac{k+py-ky^2}{(1-y)^2}$; so that the numerator of our formula must be equal to the square $k^2+2kpy+p^2y^2-2k^2y^2-2kpy^3+k^2y^4$; and, removing the second terms, by making $4k^2-8a = 2kp$, or $p = \frac{2k^2-4a}{k}$; and dividing the * Thus, $py = \frac{2eh^3}{k} \times \frac{4hk^2(2a-k^2)}{3k^4-4a^2} = \frac{8eh^4k(2a-k^2)}{3k^4-4a^2} = \frac{8k(k^2-a) \times (2a-k^2)}{3k^4-4a^2}$; also, $qy^2 = \frac{eh(k^2+2a)}{k^3} \times \frac{16h^2k^4(2a-k^2)^2}{(3k^4-4a^2)^2} = \frac{16eh^3k(k^2+2a)\times(2a-k^2)^2}{(3k^4-4a^2)^2}$ $= \frac{16k(k^2-a) \times (k^2+2a) \times (2a-k^2)^2}{(3k^4-4a^2)^2}$, by substituting $eh^4=k^2-a$. B.