OF THE FIFTH DEGREE. 135 
1 1 
the product of the roots (1 + ./2)*, (1 — ./2)°, being — 1. Put- 
ting ? for 28560, and 2 for 28562, 
72 38 
eee cE — 3, Bg eS os 
1278 RoR 
+ &x = 25, he = —*, hee 
P=2(2 + cz) — g = 5l, 
A— 8 = 
ua’ Se i ay alg 
B , 
Vv de (12 — o2 2) 48, — 518 + 14 x 338 
cat — 2 ~oo— 4)\ —— es 
B 
These values cause (22) to become 
13 _ A@ + (514A — 488)" | + 28 (1A — 48A) @ 
12 24 Q? + (5124 — 48/3)? \ + 24 (514 — 488) Q 
where Q = 484 — 518 + 14 x 338. This may be written 
13 Hi + 2Kp 
12) He + 2k 
In order that this equation may subsist, it is necessary that 
H (138 — 12) = 2K (128 — 132); 
H a 
or — 
2 2 
2 2 
But H = (— 80852)? + (85782)? = 6537045904 + 7358551524 
B—24 
= 13895597428 ; — K = (80852) (85782) = 6935646264 ; 5 
= 14268; ial — 14293 ; and 6947798714 x 14268= 6935646264 
x 14293 = 99131192051352. 
§10. Hxzample Third. I will finally take an example, 
x + 2003 + 20x? + 302 + 10 = 0, (24) 
in which the roots of the auxiliary biquadratic are all rational. By 
(6) and (10) aud (16), g = — 2,4 = —1,c?z=0. Therefore the 
denominator of the expression on the left of (22) is zero, while the 
numerator is not zero. Therefore the denominator of the expression 
on the right of (22) is zero. Or, — g* + 4k’g* — 8ek* + 4eg® k’ = 0. 
5 13\2 
5° Therefore z = =) and c = 0. Hence 
M = — 10, V = 10; and, if 
12 
Therefore e = — 
