AND IMAGINARY ROOTS OR EQUATIONS. 
623 
(42) In fact M. Hermite’s octodecimal invariant is most simply obtained as the result- 
ant of the primitive quartic and its canonizant. Using the reduced forms for these two 
but Jj, J 2 are subsequently without warning (compare expressions for AA', BB', CC', pp. 186, 192) renamed 
A , J 3 ; so that 
8J 2 =l + 16e 5 , 24J 3 = — 1 — 64e 5 . 
Tbe corresponding values of J, K, L have been already calculated, and we have found 
J=l, K=-2e 5 , L=0. 
Hence 
Thus 
A=l, g+2e 5 =B-2Ce 5 , -^-^e 3 =J)-2Z 
A=l, B=J, C= — 1, D=-^r, E=1 
o 24 6 
To find E, take another form convenient for the purpose, as a; 5 +10 dx 2 y 3 +y 3 . 
Taking the emanant of this (a?, 0, cly, clx, y'^x t , y') 4 , the quadratic covariant is obviously xy + 3d 2 y*, so that 
J=l. 
Also its discriminant is 
10 0 d 
0 0 d 0 
0 d 0 1 
y 3 -y 2 x yx? -a? 
of which the discriminant is 
viz . d 3 y 3 —d(— dx 3 + y 2 x) = d 3 y 3 — dy-x + d\v 3 , 
d w +4d 2 \ 
(w)‘= 
d 10 -^d 3 , 
Hence by definition » L=e— ^-d 10 +d 5 . 
Again, to find A, B, C, C', B', A', we must write 
x + 3d 2 y — N, 
y=Y, 
and we have then 
(X— 3cZ 2 Y) 5 + 10c?(X — 3cZ 2 Y) 2 Y 3 + Y 5 = (A, B, C, C', B', A'JX, Y) 5 . 
Since J=1 and K is of the eighth order only in the coefficients, it is obvious that neither J 3 nor JK can contain 
a term involving d 10 . In order therefore to find E, it will be sufficient to compare the coefficient of c£ lu in J 3 
and in L. 
Now A=l, B=-3 d 2 , C=9d 4 , C'=27d 5 +d, B’=81d 8 -12d 3 , A'=243cP + 90cZ 6 +l. 
Also A=J=1. Hence neglecting all but the terms which bring in d 10 , 24J 3 (p. 186, Memoir) is tantamount 
to I„, and I 2 (p. 186) is tantamount to 
which is 
2(243cZ 10 — 5 . 3 . 81cZ IO + 10 . 9 . 27 cP), 
12 x 243d 10 . 
Hence in J 3 the term containing cP is 
Hence -MF=2M or F=-18. 
4 2 7 
Hence we have, finally, 
A=J, 
J 2 =-K+1P, 
J 3 =-18L+|JK-^J 3 ; 
