290] FOR CUBIC AND QUARTIC EQUATIONS. 475 
then we have identically 
9 (3a 2 / 2 + X 2 ) ^ 2 = - 4a 2 X 3 + 36 (b 2 - ac) 3 X 2 - 4a 2 (b 2 - acf (P - 27J 2 ), 
which is impossible under the given combination of signs, since the left-hand side 
would be positive, and the right-hand side negative. 
To prove the above identity—the relation JU 3 - IU 2 H + 47T 3 + <E> 2 =0, between the 
covariants of the quartic, gives 
a 3 J + a 2 (b 2 — ac) 7 — 4 (b 2 — ac) 3 + S- 2 = 0, 
or, what is the same thing, 
But 
and thence 
or 
^ 2 = — a 3 / — a 2 (b- — ac) I + 4 (b 2 — ac) 3 . 
X = 3a/ -h 2 (¿> 2 — ac) I, 
3^ 2 + a-X = — a 2 (6 2 — ac) /+ 12 (b 2 — ac) 3 , 
3^ 2 = — a 2 X — a 2 (b 2 — ac) I + 12 (b 2 — acf, 
and the identity will be true, if 
(3X 2 + 9a 2 / 2 ) {- X - (b 2 - ac) I + 12 
(b 2 — acf} 
a 2 j 
= - 4X 3 + 36 X 2 - 4 (b 2 - acf (P - 27 J 2 ). 
This gives 
(3A r2 + 9a 2 / 2 ) {— X — (b 2 — ac) /} = — 4X 3 — 4 (6 2 — acf P, 
or, what is the same thing, 
(3X 2 + 9a 2 / 2 ) { X + (b 2 — ac) /} = 4 { X 3 + (b 2 - acf P], 
or, dividing by X + (b 2 — ac) I, 
3X 2 + 9a 2 / 2 = 4 {X 2 - X (5 2 - ac) 7 + (b 2 - acf P], 
and reducing 
X 2 — 4X (6 2 — ac) I — 9a 2 / 2 + 4 (6 2 — ac) 7 2 = 0, 
or finally 
{X — 3aJ —2 (b 2 — ac) 7} {X + 3a/ — 2 (6 2 — ac) 7} = 0, 
which is true in virtue of 
X = 3a/ + 2 (6 2 — ac) 7, 
and the identity is thus proved. 
60—2