PEOFESSOB FINLAY ON IMPOSSIBLE EQUATIONS. 241 



my-\-nz=:f (1') ; 



and by eliminating a' from equations (2') we get 



be — adrz cy' — az^ (3). 



Again, by eliminating y from equations (1) and (3), we 

 obtain 



(cti^ — am*) 2^ — 2f/icz zz: (be — ad) m^ — c/^...(A.) 

 Let Zi and z^ be the values of z deduced from this equation, 

 i/i and ^2 the corresponding values of y deduced from (1'); 

 then if each of the radicals in equation (1) be restricted to 

 a positive signification, it is evident, that if either y or z be 

 negative, the corresponding value of x will be an impossible 

 root of equation (1), and that the root of (1) corresponding 

 to t/j and Zi cannot be possible unless 1/2 and z, be both 

 positive. 



To illustrate this theory by a numerical example, let 

 the proposed equation be 



3 V(2a; -|- 5) 4. 4V (Bx — 2) n 17. 

 Comparing this with equation (1), we have 



ni=z3, n=zi, az=:2, b — 5,c=z8, dzz — 2,/= 17 ; 

 hence equations (1') and (4) become 



3y + 42r rr 17, 302^ — 408z =: — 696. 

 From the latter equation we readily obtain 



and, by substituting these in the former, we get 



Now, since y^ is negative, the corresponding value of x 

 will be an impossible root of the proposed equation; but 

 since yi and Zi are both positive, we see that the equation 

 has a possible root. To find the possible root, let z zzzi 

 rz 2 in the second of equations (2) ; then 



^(Sx — . 2) =: 2, and .-. a; =: 2, 

 which will be found to satisfy the proposed equation. 



To find the impossible root, letz =: z^zz: «/ in the same 

 equation; then 



V(3a: — 2) = V» and .-. x = 3^*, 



2 I 



