530 
PROFESSOR CAYLEY’S EIGHTH MEMOIR ON QUANTICS. 
(x, y ), assumed to lie in the region P or Q, will lie in the one or the other region 
according as it lies on the one side or the other side of the line in question, viz. in the 
region P if x-\-[hy is = — , in the region Q if x-\-y>y is = + . But we have 
2 n L— J 3 + |uJH 
J 3 
and J being by supposition negative, the sign of 2 n L — P-j-^JD is opposite to that of 
x-\-yjy. The region is thus P or Q according to the sign of 2 n L— P-j-^JD ; and com- 
pleting the enunciation, we have, finally, the following criteria for the number of real 
roots of a given quintic equation, viz. 
If D= — , the character is 3r+2 i. 
If D= + , J=+, then it is 
But if D=+, J= — , then being any number at pleasure between the limits -\-l and 
— 2, both inclusive, if 
2 n L— P+|«JD=+, the character is 5 r, 
2 I1 L— P+ P JD= — , „ „ „ r+4i. 
284. The characters 5 r of the region P and r-\-4i of the regions Q and T may be 
ascertained by means of the equation (a, 0, c, 0, e, 0]$A, 1) 5 =0, that is 
0(a0*+lOcd 2 +5e)=O; 
there is always the real root 4=0, and the equation will thus have the character 5 r or 
r-\-4i according as the reduced equation a0*-4-lOcO‘ 2 -\-5e=Q has the character 4r or 4 i. 
It is clear that (a, e) must have the same sign, for otherwise 4 2 would have two real 
values, one positive, the other negative, and the character would be 2r-|-2«. And 
(a, e) having the same sign, then the character will be 4 r, if 6 2 has two real positive 
values, that is, if ae — 5 c 2 is = — , and the sign of c be opposite to that of a and e, or, 
what is the same thing, if ce be = — ; but if these two conditions are not satisfied, then 
the values of Q 2 will be imaginary, or else real and negative, and in either case the 
character will be 4 r. 
285. Now, for the equation in question, putting in the Tables b=d=f= 0, we find 
D=256 ae 3 (ae-5c 2 ) 2 , 
J = 16 ce (ae+Sc 2 ), 
2“L— P= 2 12 ce 3 {2(ae- cy-^ae+Sc 3 ) 3 } 
= 2 12 ce\ae— 5c 2 )(2aV+a 2 cV+8ac 4 ^+5c 6 ). 
We have by supposition D= + > that is, ae= + ; hence J has the same sign as ce ; whence 
if J= + , then also ce=-\~, and the character is 4 i; that is the character of the region 
T is r-j-4i But if J=— , then also ce— — . But ae being =+, the sign of 2 n L— P 
is the same as that of ce(ae — 5c 2 ), and therefore the opposite of that of ae — 5c 2 : hence 
D = + , J=— , the quartic equation has the character 4 r or 4 i according as 2 n L— P is 
= -(- or = — . Whence the region P has the character 5 r and the region Q the 
character r-\-4i\ and the demonstration is thus completed. 
