274 Wiider’s Algebraic Solution. 
2p=—a, (1) 
2q=b, (2) 
x4 -+3p2 =2e, (3) 
q=4, (4) 
= Se 
‘6 4 = +167 Fs i624 ©? (7); 
or better, 2'* —2p2x°+-p* —4q*p — —q'=0: 
eliminating «* from (3), (5) and (7) by the process indicated, 
and we have 
metal p?+q'=pcte, (8) 
q(3)— (6) pa=a+f, (9) 
p(6)—4(5) q=ge—f- (10). 
_ The equations (8), (9) and (10) are satisfied by making p? =¢ 
and g?=e, f being equal to nothing. This changes the 
given equation, by writing —2p for a, and —2q for —b, into 
y? —2py*—2qy> +p*y! +2pqy? +9°y? +3 =0, or better, 
ne ee ae Se 
which shows that. the reduced is nothing but the rule of 
Des Cartes, applied to the above equation. 
Since the given is parted into two factors, 
(y* —py? —qy+V g)y' —py* qd —Vg¢)=0, the rule ap- 
plied to y* —py? — qy+/V g=0, gives for the reduced 
| (a), 294 2pxt +p?x? —p?=—AV £5 changing the 
signs of the second and fourth term of the first number, 
we have (b), x* — 2px* px? -+q? = —4vV ¢ These two 
equations give x212—2p?x'+(p*+4pq?)2*—q'=168; 
which is (7). It is easily seen that we have another equa 
tion, (e*—p*)y*—py? — qy=(e*—p*) g, which is the 
same as the given equation. 
a £2 XG ; 
- We might have treated the function m otherwise by 
writing it thus, 
x1# +(8y* — Spy? —8qy ~ 2p?) 
at a ee aya Ci = peg 
(Gay*— 4py?—4qytp*—4pq")x'—g* G, 
8 4 Qyx? +(2y? —p)a+q . a’ and when(e)=9, 
we have by transposition, and extracting the square root, and 
