88 4 New Method of Resolving certain Equations. 
This solution fails (in the case where no solution is 
needed) when a?=6, For then the expression (B) be- 
comes z= Be Se whence <= —a; which would be, in 
most cases, false, the true werae of x oe (-a+*/a? —c). 
To explain this, let it be observed, that the assumption 
(a'c'=b'?), see (A), requires that =< whenever 6=a?. 
(E 
When the lower term of (B) vanishes, the upper van- 
ishes with it 
For ser (4 3fb—a*.rta,.. rta?+b—a*-r+a 
=o; Or, r+a*’ +hb—a?=r?+2ar+b=0. 
o 
Phssetore, z== —~ 5 an expression which might seem in- 
determinate. In reality, however, since both terms vanish 
together, and the upper is in its form of two dimensions, 
while the lower is only of one, the whole ee will 
1, and z=—o0; o=r. 
To investigate the cases in which this will occur ; 
Since (4/? =a'c’) and b' is now equal to nothing, either Sor or 
c’ must equal nothing. But ifa’ is 0, & —a@’2=0. ~. 72+ 
Qar+b —(r? + 2ar+a? )=b ~— a? =0, which cannot be sup- 
posed, for the reason stated in (E). 
Next suppose c’ to become evanescent. bi 
r°-+-3ar? +-3br-+-e =0 
But (r?+2ar+6) (r+a)= Seba 4. MESES 
Whence 2) —a?.r-+-c—ab=0 
This therefore is the case already considered in (C). 
