230 ME. R. hahley on impossible 



Let us recur to the examples cited by Mr. Cockle from 

 the Vija-ganita. And first, as it regards the bee question, 

 if, as directed, we denote the number required by 2^% the 

 conditions of the question will be expressed by the equa- 

 tion 



X+-Q ^ + 2 = 231^^ 



which gives x zr 6, — the negative root being rejected ; and 

 consequently 2^7* 1:1: 72, the true and only answer. 



But if x (instead of 2a!^) be taken as the qiccesitum of the 

 question, the equation will be 



ViaJ-j- I a? + 2 = x; 



or, 2a; — 9 V2x rr 36 (A); 



whence, by the usual process, we find x zz 72 or |, the 

 latter root being foreign, and belonging to the congener of 

 (A), viz., _ 



2a; + 9 ^2x = 86 (B), 



and the former root being alone the true answer. 



Again; to solve the second example from the Vija- 

 ganitay let us assume, as instructed, a' for the whole number 

 of arrows; then we shall have 



I- s^ -\- 4x -}- 10 =z x' ', 

 whence « zz 10, — the negative root being rejected as 

 before ; and /. «* zz 100, the number required. 



But taking x (instead of a;^) for the number sought, we 

 shall have 



|a;+ 4 Vx_-\- 10 — z; 



.\x — 8 V« = 10 (C); 



whence, resolving as usual, we get x zz 100, or 4. The 

 former root satisfies (C), and is therefore the answer to the 

 question; the latter belongs to the congener of (C), viz., 



x-{-S \^x=z 10 (D). 



In passing, X may just observe that (A, B), (C, D), 

 being put under the forms — 



(2x — 36) + 9 V2 a; — 36) -f- 36 n 0, 

 ( a; — 10) + 8 V( a; — 10) -f 10 = 0, 



