Wilder’s Algebraic Solution. 279 
A=3p, (1), 
=3q+3Sp? — 4d, (2), 
C= 6pq+p* — 3ap,(3), 
3aq-+3p? aoe -aphinhae 4+: 3p3q 4), 
6apq+p*a—3pa? —2ab+ ee be =3pq?, (5) 
2? +-9*=6pqb—3pab—b?, 
If we malsiall (4) by p, and subtract it seat (5), we shall 
have (3q —2a)p* ray. A ieee ne 2a)b=0. ‘his equa- 
tion is satisfied by ma 
2 
Writing 5 q @ for q, in on and (6,) and we have, 
ap? +3bp——- =0; an8 
=bp® +-abp —b?— em ; from whence 
we have y ; oe when y?-+ay+b=0, we have y?+py+q 
—x*=0; and we have already p, q and 2, in functions a and 
b. This is the rule of Tschirnaus. We may still vary the 
calculation, by assuming the function, 
(3 +0? + (a+p)a+b+9+a)(8°+8? + ey 
(72+7? +(a+p)y+b+q+2) ; 
for, making the coefficients of x and x?, equal to nothing, 
and eliminating the symmetrical function «+-8-+-y, «7+? 
+y?, etc. by means of aand b, we obtain, 
Se: es 
ap! + 3bp —> =0, and 
=bp?--abp—b? —5=-° 
now, if we ae y?+ay+b=0, we aan have y? +py+4¢ 
=0, 2s whence y is known, y standing for one “Of e let- 
ters a, ry. 
Let as sat recall the function, 
xt —(y* — 4py* +-4py +2p*)x* +(p* — 4ayp* + Aq’ pt 
eek alk ack ©, 
2q7y?)x*-+-q' : (G) 
(H) 
