134 RESOLUTION OF SOLVABLE EQUATIONS 
aa 89 (° x 412)) 
12 x 892 4] 22 
where h = B x41 x (RY = 39 V and é= — 39 a 
We have to show that this is the result to which the equations of the 
preceding section lead. The simplest way will be to find g, & and cz 
by means of (6), (10) and (16), and then to take the values of e and 
./z given above, and to substitute them in equation (22). If the 
theory is sound, the equation ought in this way to be satisfied. When 
this equation has been satisfied, it will be unnecessary to pursue the 
verification farther. Because 
11 vt da 7a 11 
22 ee 
Pp = Sr and ps = — 95» I = oR 20 x 95° 
5 
From (18), taken in connection with (21), che must be negative. 
Therefore 
ae cdeadl ls, * SERA ge on 8 =) 
4x 3) ee 80 x 41 \25 7’ 
hes = — 15 "tay (G)# -— ** = — G05 
yf 2 / 2 
= — aim (a5): ” = a im Gs), 
M — 2e (2? — 2) = — = (35): 
These values reduce the equation (22) to the identity 
89 89 153 (40802 + 5 x 12242) — 89 (2448 x 530) f 
Ag: = et ECO TL nad . 
89. Example Second. The example that has been given is one in 
which the auxiliary biquadratic is irreducible, I will now take an 
example, 
x + 1023 — 8002 + 145” — 480 = 0, (23) 
in which the auxiliary biquadratic has a sub-auxiliary quadratic. 
When the root of the equation (23) is put in the form (1), 
1 a 
uw = (1+ 2), ut = (1 — V2)”, 
2 
5 
ue = (1 + 2) (1 + 42), 
ug = (1 — /2)(1 ors 2), 
