^14 MR. B. HARLEY ON IMPOSSIBLE 



recur to the first introduction of the symbol n into the 

 equation (a), we find that it was employed as a multiplier 



of a negative quantity ( — V 2aa; -f- b), in order that that 



quantity might be made to assume a positive form. Hence, 

 therefore, in the foregoing transposition of b n, we must 



multiply (and not divide) it by n. So, in like manner, in 



extracting the root of the quadratic, half the co-efficient of 



JB is muUipUed by n, and the square of the result is then 



I 



added to the absolute term. Had we eliminated V from 



(a), by multiplying by a) -] — V 2ax -f- b, instead of ;c -j- ** 



V* 2a^ -}■ b, we should then have had to divide every trans^ 



111 

 posed quantity by n, or (which is the same thing) to mul^ 



tiply it by n-^ : the value of x so evolved would be found to 

 differ from that above given only in form. In fact, since 

 «p in — 1, JO being any odd number, positive or negative, 

 yffe might (if we chose) employ n" for n, as a midtiplier, 

 throughout the investigation. Thus conducted, the opera- 

 tion would be as below : — 



Multiplying (a) bv ^ + n^ *j2,cuc -4- ^> and bearing in 

 1 " 1 11 1 



mind that 1 + ^^ ^= 0, we have 



x" -{- nP (2a X -\- b) z=i Q> ; 

 \ 11 1 



.', a? '\- ta nP X :zi b n^ ; 



.-. a; =: o n^ 4- V«* «*» -4- b »'p 



11 1 1 



11 II 



It is easily seen that, since p is odd^ (1) and (P) are vir- 

 tually identical. 



We now proceed to verify these solutions — 

 First, by (1,) V2aa;-j-6— -v/{2oV-|-2a» VoV + i+i} 



=: V {(aW -}-i) -f2ff« V«V+6-f aV) 



•=L »^a^n*-\-b -fan (2) 



1 1 1 



