136 RESOLUTION OF SOLVABLE EQUATIONS 
D = M*? — zN* — 4e? (kh? — c?2z)?, 
D = — 104e?. Therefore, by (21), @= — a 
= 0 2 : Therefore by (9), A a 
=o gist erefore by (9), h = 6° 
Therefore using the symbols, B, B’, as in §3, 
5 45 
5 9g = A+ V2) = 81, 
% = h (2 —/z) = 36. 
Therefore wW=—749=2,u=—7—9= — 16, 
Hence, by (3), 
which is the solution of the equation (24). 
§11. It was pointed out in §7 that, in the case we are considering, 
there are six equations and five unknown quantities. All the 
unknown quantities may be eliminated, and an equation p’ = 0 
obtained ; where p’ is a rational function of the coefficients po, pz, 
etc. This elimination has been performed, under the direction of the 
author of the paper, by Mr. Warren Reid of Toronto, with the 
following result. Putting P, as in $7, for 2 (k? + ¢? z) — g’, let 
A = — 2ke? 29° 38 (2? + o? 2) — 39° }, 
i § 16K? c? 2 + 4 (A? + c? 2)? — 598 (kK? 4+ c? 2) 4 g® i, 
D= — 4(h% — cz) i + 398 (k? + c? z) — 2 (k? — cz)? ls 
A= — 8ke?2[32ke?2(k?—c*z)—P | psg? + 8k(k?— cz) — 4hg® i] 
a 
} psg?+ 8k(k?—c?z) —4hg8 [—32470?2-+ 9° | 4(k?+ 0°72) —g8 ‘| 
+ 64kc? z P (k? — c* 2), 
D,= — 16he? 298 {4 (k? + oz) —g® } 
+ 4P (? — 0? 2) {ps g? + 8k (hk? — c? z) — 4kg i. 
Then, since 109 = — po, and 20k = — ps, and 
20c2 2 = pag — 5g? + 20K?, 
