Elements of Algebra

CHAP. IX. OF ALGEBRA. 373 $q = \frac{6eh^2-p^2}{2k} = \frac{3eh^4k^2-2e^2h^6*}{k^3} = \frac{eh^4(3k^2-2eh^4)}{k^3}$; or, lastly, $q = \frac{eh^2(k^2+2a)\dagger}{k^3}$, because $eh^4 = k^2-a$; after which, the remaining terms, $4ehy^3 + ey^4$, being divided by $y^3$, will give $4eh + ey = 2pq + q^2y$, whence we find $y = \frac{4eh-2pq}{q^2-e}$; and the numerator of this fraction may be thrown into the form $\frac{4ehk^4-4e^2h^5(k^2+2a)\ddagger}{k^4}$, or, because $eh^4 = k^2-a$, into this, $\frac{4ehk^4-4eh(k^2-a) \times (k^2+2a)}{k^4} = \frac{4eh(-ak^2+2a^2)}{k^4} = \frac{4aeh(2a-k^2)}{k^4}$. With regard to the denominator $q^2-e$, since $q = \frac{eh^2(k^2+2a)}{k^3}$, and $eh^4 = k^2-a$, it becomes $\frac{e(k^2-a) \times (k^2+2a)^2-ek^6}{k^6} = \frac{e(3ak^4-4a^3)}{k^6} = \frac{ea(3k^4-4a^2)}{k^6}$, so that the value sought will be $y = \frac{4aeh(2a-k^2)}{k^4} \times \frac{k^6}{ae(3k^4-4a^2)}$, or, $y = \frac{4hk^2(2a-k^2)}{3k^4-4a^2}$; and, consequently, $x = y+h = \frac{h(8ak^2-k^4-4a^2)}{3k^4-4a^2}$, or $x = \frac{h(k^4-8ak^2+4a^2)}{4a^2-3k^4}$. * By multiplying $6eh^2-p^2$ by $k^2$, and substituting for $k^2p^2$ its equal, $2eh^5$. † For since $k^2 = a+eh^4$, therefore $3k^2-2eh^4 = 3a+eh^4 = k^2+2a$. ‡ Here $4eh = \frac{4ehk^4}{k^4}$, also $q = \frac{eh^2(k^2+2a)}{k^3}$, and $p = \frac{2eh^3}{k}$; therefore $2pq = \frac{4e^2h^5(k^2+2a)}{k^4}$, and, consequently, $4eh-2pq = \frac{4ehk^4-4e^2h^5(k^2+2a)}{k^4}$. B.