232 MR. R. HAELEY ON IMPOSSIBLE 



remaining corresponding equations, viz., 



4 _ ^x — 3 + Vic + 21 :=z 0, 

 and 4 — a/x — 3 — V^^ 4- 21 — 0, 

 being also impossible. 



Again, let the equation 



Va;* + 9 + V25— x^ — 2x=z0 

 be proposed. Assume Va:" -^9 =i o!, and V25 — a?* zz a;; 



1 s 



then 



a: -4- a; — 2x m 0, 



1 2 



a^ 4. ar" — 84 rz 0, 



1 2 



ar^ — x + 9—0. 



Eliminating j? between the first and third of these equa- 



tionS) we get 



(x + ar)* — 4 (.r^ — 9) zr ; 

 1 2 1 



whence, by means of the second equation, we obtain 

 a: 3: 5, — 5, -^r, or :^ ; 



and K =: 3, — 3, — - ii, or 14. 



Now, since no corresponding pair of these roots, except 

 the first, is positive, there is only one possible value 

 of a? capable of satisfying the proposed equation, viz., 



(cc ~i~ ^ \ 

 A !=::] 4. The information afforded by the signs of 



the other roots is, that the equation 



Va^ + 9 — ^25 — 3^-{'2x=zO 

 has one root, and one only, viz., 2_ and that the equation 



V5, 



^x» 4- 9 + ^25 — a^ — 2a: = 

 is impossible. 



To recur to the equation 



Va 4- i^x-{-l = 0, 



