MR. MURPHY ON THE ANALYSIS OF THE ROOTS OF EQUATIONS. 
171 
x, = x 3 = x 4 , therefore the equation <p (x) = 0 has three equal roots, and since a, 1 " = 0 
denotes the existence of two equal roots, therefore a" = 0 is the additional condition 
for a third, or the system of equations a" = 0, a'" = 0, are equivalent to the system 
<p (x) = 0, <p' (<r) = 0, <p" (x) = 0. 
lx') 
Now — — = 6 x 2 + 3 a x -f b = z -f- b for abridgment, we shall next form an equa- 
tion in 2 indicative of two equal roots, and eliminating z by the equation z -j- b = 0 
we shall obtain a". 
When x x = x 2 , then 6 x 4 2 -j- 3 a x x — 6 x 4 2 — 3 x 4 (2 x 4 -|- x 3 -f- x 4 ) 
= — 3 (x 4 x 3 + x 4 x 4 ) 
= — 3 (x 4 x 3 + x 2 x 4 ) 
the equation 
x + 3 (x 4 x 3 + x 2 x 4 ) = 0 
expresses that x 4 = x 2 , and forming a cubic equation by taking all the symmetrical 
simple equations, a condition in z for the existence of any pair of equal roots is ob- 
tained, viz. 
F (a) = {z + 3 (x 4 x 3 x 2 x 4 )} . {z + 3 fa x 2 + x 3 x 4 )} . { % + 3 (x 4 x 4 + x 2 x 3 ) } = 0 
or 
F (z) = z 3 -}- 3 b z 2 -f- 9 (a c — 4 d) z -f- 27 {d (a 2 — 4 b) -J- e 2 } = 0 ; 
and if z be eliminated between this equation, and z -j- b — 0, the result multiplied by 
a constant will be a! 1 or 
a" = h’ F(- b). 
Now one factor of 
F ( — b) — 2 (x 4 x 3 + x 2 x 4 ) — (x 4 x 2 -{- x 4 x 4 -f- x 2 x 3 + x 3 x 4 ) 
= fa — a? 2 ) ( x ?< ~ x d + ( x i - x d Os ~ x 2 ) ; 
the other two being symmetrical with it gives the same value of a" as before, and 
h! = Jt. 
31. If al — 0, a" = 0, a!" = 0 simultaneously, then x 4 = x 2 = x 3 = x 4 . Now «' is 
a function of two dimensions, as is that arising by eliminating x between <p" (x) = 0, 
<p m (x) = 0, and gives the same condition, but <p"’ (x) = 6 (4 x + a), therefore^" 
is the result of this elimination, therefore a! = h" <p" Now 
<p" (x) = 2 (6 x 2 -J- 3 a x + b) 
therefore 
n ( “ a \ ~ 7 3 a 2 
and our former value of ot is 
k" (2 3 X] 2 — 2 2 x x x 2 } = k" (3 a 2 — 8 b ) 
h!' = - 4 k". 
hence 
