130 RESOLUTION OF SOLVABLE EQUATIONS 
wh = ES et evet Ot evavs} 
{k—cVz+ (9-9 V%ASH} 
=A+t+A f/2+4+ (A° 4+ A" J/2) V8, 
where A, 4’, A”, A’”, are rational. The value of A is 
A = —aaalg (# — 8) + aes (02 — g?2)t. (11) 
Bis nate 2 
eae i ae That is, from (8) and (5) and (7). 
2 43 
2 
a ees L2btevx—G—ate) Gt avs 
+2(kKteVY2IteyYz)v/st 
jk—cVYz+ (9-9 V2)Va} 
=B+BY2z+ (PtP J/2aVvs; 
where B, B’, B’, B’”’, are rational. Now, by (4), 
S, = 20 1% (uf ue ) + 3% (82) + 60 uy wg ug ug. 
And S, = 2p2 — 4py. Also 2% (ui uw) = 44; and, by (6), 
10g = — pz; and, by (5), um wz ws ws = go — a, %. Therefore 
pa = — 204A + 5g? 4+ 1da? z. (12) 
Again, Ss = 5 {y (ui) } + 2 (S283) + 50 (2 (mu us ui) 
And ¥ (uj) = 4B, Sz Ss = 6 pz pg = 1200 gh, and 
2 (um us us) = Uy U4 (uz uw + 3 us) + un us (uj us + us U2 ): 
Therefore S; = 206 + 1000gk — 200acz. 
But Ss — 5p2 ps + dps = Ss — 1000gk + 5p5 = 0. 
Therefore ps = — 4B + 40acz. | (13) 
The values of pg and ps in (12) and (13) make thé quintic 
F (x) = @ + po a8 + pga? + (5g? + 15022 — 20A) x 
+ (40acz — 4B) = 0. (14) 
§4. Assuming the coefficients p2, p3, etc., in (1), to be known, 
the coefficients in the equation / (x) = 0 as exhibited in (14) involve 
six unknown quantities, namely, a, c, 0, ¢, e, h. The list does not 
