62 G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=0. 
Then we have: . 
gaa ; 
(49) A =—57(P-E+Q) - AE). 
The conditions that F(e’)=0 shall be satisfied by three real and positive 
roots é*, and consequently that there should be three real values of e for a given 
value of E are thus: 
(a) M,= —P’—3E>0 
(b) M,=3P?—{Q'E>0 
2s Ae: 
(c) A =—(PE+Q)HE>0. 
The discussion of these inequalities necessitates that we consider the roots 
of the equation . 
A,h=0.~ 
We know that they are all three real by the very fact that [> 0. 
Let h,, h., h;, be the three roots. The comparison of 7,(/) with its expres- 
sion in function of the roots gives : 
pe =0 ,namely, h+h; +h; =0 
(50) La =F Ayha + hzh, + hoh3 =P’ 
— hyhihs = Q’ e 
The first equality, 24,=0, shows that one at least of the roots must be 
positive, and one at least must be negative. We fix the notation in accordance 
with the hypothesis : 
h, > he>-hy. 
This comes to the admission h, >0, h;<0. As to the sign of h, it is the same 
as that of Q’, which is supposed a datum, because @’ may be put under the 
form Q’=h,(—/;) x h, and h,(—h,) is positive. 
For the moment we put : 
(50 bis) hz=hz, then h,=—h,(1+2). 
Applying to these values the hypothesis h,>h,>h3, we get for z the limits 
1>2>—5- 
As by squaring 22,=0 we get 
Thyh=—5 2M, we have 
P=- ; h[1+2°+(1+z2)*], namely 
