134 
- illA±Ax^ + ~ ^(«/3 + gy /3v) ^ ^ .^7) 
m* ’ ’ \ / 
Now this equation will coincide with x^ + aa^ + hx + c = 0 
when 
a + P + c 
m^a 
n ri VtlHc^ - 4c) 
aj3 + ay + Qv = — - 
lo 
a(^y 
~U 
Hence, a, /3, v are the three roots of the auxiliary cubic 
o m^a „ mHa^ - 4c) ^ 
2“ + — 2^ -1 ^ “ = 0 
2 16 64 
(28) 
As this auxiliary is the same when (h) is negative, it will 
be necessary to change the signs of so as to 
make the coefficient of x in (27) negative — this is easily 
done when required. This mode of procedure renders 
unnecessary the solution of two quadratics. 
7. The following, which has occurred to me, seems to 
combine all the advantages of the forms (B) and (C). 
Assume — 
{i(^ + a.x ^ (3 + V i'} ~ + \/ (3 - \/^y} =0...(29) 
The biquadratic formed by multiplying out. 
+ [2-^ (3 — n^a)x^ — 2nyJ ay.x + j3 — v = 0 ( 30 ) 
The coincidence of (30) with (2) gi ves 
2-)/ (3- n^a = a 
— 2n \J av — ±5 y ( 31 ) 
(3 — y = C J 
From these conditions it follows that 
b 
V'v = =F; 
VP = 
With an auxiliary cubic 
2n^/a 
a + n^a 
a^ + 
2a 
n 
a^ + 
- 4c 6^ 
= 0. 
or 
Let r, s, t, be the three roots of this auxiliary cubic. 
.( 32 ) 
