G. PLARR ON THE SOLUTION OF THE EQUATION VpGp=0 3 61 
8, The question reduces itself to the determination of the limits between 
which E must be ney in order that F(¢)=0 should have three Borne 
roots. 
, — the solution. of this question, we ha Fe? with 
F(e’) =e —3M,¢ + Me’? — la 
MM, OP ei A= eee 
where i: 
| M,=3P?—QE, 
and calculating . Pe 
X=F(M;) , Y=F(M,) 
A= —4X0— Di Ne 
we have to put the condition : 
A>0. 
From (46) we deduce, by differentiation : | 
Fe =3( +P) +5 B(Ee —Q'‘); 
This gives 
hg 26s a Ge} be 
| X=pE'—5E (qE" + aes +Q) 
and by (46) : 
| Y= —geE* + po( gE? + EP’ +o). 
‘We introduce : i , Q+EP’=u, only for the moment. 
Be tikiting and reducing, we get: 
X= ape uw}. 
ae 33 
aye ov + opuut oeu 

Then as E’=v, we deduce: oy 
SAS =2 "of oe Vv oe gq U ~u | = oY. 5 
Developing and ome the terms depending on vt, va, vu, disappear, 
and there remains : am 
a oo 39 Ps : 
A=— oi (v+u) : 
But v+u=E? + P’E+Q’; this is what the faction Ih=/ (m+), ae, 
Ap=hP+Ph+Q>°~ 
becomes for «=0 and b= E. We put 
(s)= Gh= =U EP RE Q* 
VOL. XXVIII. PART I. Q 
