238 PBOFESSOB FINLAY OK IMPOSSIBLE HQUATIONS. 



»/(a — 5V — 1); for if we took — a-f-/3 V — 1, when/3zr 0, 

 we should have — a for the arithmetical value of V a, vvhich 

 would be inconsistent with our original assumptions and 

 restrictions. 



IL 



Let us now consider the equation 



X -^ m *^(2ax -\- b) zn c (1), 



where w, a, b, c, denote given numbers, which may be 

 positive or negative, fractional or entire. If we assume 



V(2«x + i)=:y (2), 



and therefore 



2ax -^ b :zz t/*, or X zz ^—5 — , 

 the proposed equation becomes 



-2^ \-myz:zc,ov 



1^ ^ 2amy =. 2ac -\- b (3)! 



Solving this equation by the ordinary rule for quadratics, 

 we get 



y zz.-^ am -\- »^{a^m^ -|" 2ac -j- ^>) or 



y = — am -j-B (4); 



where, for the sake of brevity, we use R to denote the 

 arithmetical square root of the quantity aW -{- 2ac + i- 



Now, if the radical in equation (1) be restricted to its 

 arithmetical value, it is evident from (2) that y must be 

 positive; and therefore all negative values oft/ must be re- 

 jected as leading to values of x, which are impossible- Thus 

 we see, that the roots of (1) will be hoth possible when the 

 roots of (3) are both positive, and both impossible when 

 the roots of (3) are both negative; but when one of the 

 roots of (3) is positive and the other negative, one of the 

 roots of (1) will be possible and the other impossible. 



Firstf Let m and a have the same signs; then — ma is 

 negative, and the lower sign must be rejected in equation 

 (4), as giving a value of y essentially negative. 



