AND IMAGINARY ROOTS OF EQUATIONS. 
589 
that all the terms free from a andy contain b 2 e 2 as a factor ( 14 ). Hence, in fact, the 
discriminant will be only of the twelfth degree in g, rj, and being therefore of only the 
second degree in g 5 +^ 5 , will admit of comparatively easy treatment. 
(7) Before proceeding to the calculation of this discriminant, it will be useful to 
investigate, as a Lemma ancillary to the subsequent discussion, under what conditions 
four of the roots of the supposed equation will become imaginary when g =>7. 
In this case writing \- V ~=-z , the equation 
^y(l, g, g 2 , g 2 , g, 1X^, ^) 5 =° 
becomes 
Z 2 — 2 — Z + l + 5g(2 — l) + 10g 2 =;z 2 -|-(5g — l)z-j-10g 2 — 5g — 1 = 0, 
or say/z=0. 
The determinant of f{z) is thus (5g — l) 2 — 40g 2 + 20g+y, i. e. 5(1— g)(l + 3g) ; and all 
the roots of z, and consequently of (x, y), will be impossible, unless z lies between 
1 and — ij. 
Now /(2)=1 +5g+10g 2 , 
/(2)=3+5«; 
so that when z has any real roots, i. e. when g lies between 1 and — /(2), f(2) are 
both positive, and the Sturmian functions are of the signs + + +• 
Again, 
/(_2)=5-15g + 10g 2 =5(l-g)(l-2g), 
/(-2)=-6+5«; 
so that, on the same supposition as before, the Sturmian functions are + — j- , viz. 
H b when |>g> — if, 
1- when l>g>^. 
In the former case two real roots, in the latter one real root of z lies between 2, —2. 
Hence in the former case no real roots of z lie between the limits 00, 2, and the limits 
—2, — 00, and in the latter case one real root lies between those limits. Hence x, y 
will have four imaginary roots, unless g lies between 1 and and two such roots in 
every other case. 
Thus the discriminant of (1, g, g 2 , f, q, Y$x, y) 5 , when g=j?, is negative when g lies be- 
tween 1 and but for every other value of g is positive, save that it vanishes when 
g=l, or s=^( 15 ), or g=— J. 
(8) I now proceed to calculate the discriminant of the form 
x s + 5g x 4 y + 10g V^ 2 + 1 0 jfafy 3 + 5 qxy* -\-y 5 
( 14 ) For the discriminant of i/)=the discriminant of <p(as, y ) multiphed hy the square of the product of 
the resultant of ( x , <p) and of (y, <p). 
( l6 ) When e=g the discriminant of f(z) does not vanish, but z= —2 satisfies the equation in z, and con- 
sequently <2 has two equal roots — 1, so that the discriminant of the original equation vanishes. 
y 4 e2 
