88 A JVezo Method of Resolving certain Equations » 



This solution fails (in the case where no solution is 

 needed) when a^=b. For then the expression (B) be- 

 comes z= — r+a, whence a;= —a; which would be, in 

 most cases, false, the true value of a; being (-aH- ^\/a^ ~c). 



To explain this, let it be observed, that the assumption 

 (a'c'=6'^), see (A), requires that c=a^, whenever 6=a^. 



(E) 



When the lower term of (B) vanishes, the upper van- 

 ishes with it. 



For as r-{-a= — ^\/ b-a^.r-i- a, .'. r+a'^+b-a^'r-^-a 

 = 0- Or, r-\-a -\-b — a^=r'^-{-2ar-\-b=^o. 



... 



Therefore, z=^ — - ; an expression which might seem in- 

 determinate. Tn 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 expression will 

 vanish, and z== — o; x=r. 



To investigate the cases in which this will occur ; — 

 Since [b'^^=ac') and b' is now equal to nothing, either a' or 

 c' must equal nothing. But ifa' is 0, 6' — a'2=0. .'.r^-f 

 Qar-\-b — {r^-\-'2'ar-\-a")=b — a^^=0, which cannot be sup- 

 posed, for the reason stated in (E). 



Next suppose c' to become evanescent. Or, 



r^-\-Sar^-\-3br-\-c =0. 



But (r2+2ar-f6)(r+a)= r=^ + 3qr^+ br-^2a-r-\-ab—Q. 



Whence 2b—a^.r-\-c — ab~0 



This therefore is the case already considered in (C). 



