282 Wilder’s Algebraic Solution. 
king (D)=0, then (C)=0; treating p as unknown, and com- 
paring with p?+-bp+c=0; we have 
3 
v3y=—b, (1), 
w P08 FC, (2), 
p=—(«?+yz), (D). ; 
From (1) and (2) we have «12 —cr®— 55 =0; 
b . (ae 
from (1) then y=— 375 3 these two equations joined to (D), 
determine p. 
Let us take for the last example 
y° +by* +ey? dy? +cy-+f=0. | 
There are several functions, resulting from different values 
of m, equally proper to resolve this equation. The one, in 
which the function is most easily calculated, is that in which 
m==2. ‘Our function is then, 
oe! 490° +S ,2°+8,21+8,27+S8, , (A’). 
eo --yet+prt+qr?+re+s “aeB') 
¥ =P 
a for 
Caleulating S,,S,, ete., and afterwards writing 
as 
6 or q, 
p, and ° we shall have, 
—y® 
9 B Brit 
—y qy 
+6py? | 1 —9p2y* | 1 +sy> Ut 
a? — px? + 8qy (aur tery" 36° + Spsy gts? (4) 
ney —12pqy +298 
= ite 
+24r J — 72sy 
wn 4g? | 
—3pr J 
=) Se a yee 
Mine sgl SSeS 6 0), +ra+s (B) 
_ Now, if we make (B’)=0, (A’) will also be equal to noth- 
ing; and the five independents 2, p,r, q, and s, allowing 
as many separate hypotheses ; we therefore make y, being 
the es 6 
same in (Ry: and y°+-by* +-cy? +-dy? +ey+f=0, 
