578 MR TALBOT ON SOME MATHEMATICAL RESEARCHES. 



This conclusion is obviously of great importance, since it shows that we have 

 only to compute the value of </> in the given equation x 3 + qx — r = o, and we 

 obtain <p = R\/b, where R is some rational quantity. The unknown quantity 

 which we have called b separates itself naturally, so to speak, and comes to light, 

 being left by itself under the radical sign. 



And, once that b is known, a is easily found from the equation q = — (3a 2 + b). 

 And, therefore, the three roots of the equation a + Vb, a — Vb, and — 2a become 

 known. 



I will now proceed to apply this theory to some numerical examples, which 

 will make it perfectly clear. 



I will first observe that the solutions obtained resemble, in one respect at 

 least, the theorem given in most treatises of algebra. "If an equation has two 

 equal roots they can be found." You do not perceive from the aspect of the 

 equation that it has two equal roots, but you are desired to make use of a certain 

 process, and then the equal roots (if any) will appear. So here it is not apparent 

 that an equation has roots of the form a + Vb, but if it has, and \/b be in its 

 lowest terms, both a and b can be found. 



Examples of this mode of solution. 



Example 1. Let x 3 — 16# — 24 = 0. Here q = — 16, r = 24. We compute 



<p as follows : — 



- q z = 4096 .-. - 4tq 3 = 16384 

 r 2 = 576 .-. 27 r* = 15552 



.-. </) 2 = - 4q 3 - 27r 2 = 832 



And (p will be V832, which we must reduce to its lowest terms. We find that 

 832 is divisible 6 times by 2, the successive quotients being 416, 208, 104, 52, 26, 

 13. Hence 4> 2 = 2 6 . 13 and </> = 2V12T The number 13 left under the radical 

 sign is therefore = b. To find a we proceed as follows : b — 13, q = — 16 

 .-. b + q = — 3, but this is equal to — 3a 2 .-. a = 1. Therefore the roots of the 

 equation are 1 + \/13, 1 — \/13 and — 2. 



Example 2. Let the equation be x 3 — 105# — 50 = 0. Here q — — 105, 

 r = 50 . Therefore — 4q s — 27r 2 or (p 2 comes out 4563000. This is divisible 

 by 10 3 , the quotient is 4563 which is 3 times divisible by 3, the quotients 

 being 1521, 507, 169, and the last number is seen to be the square of 13. Hence 

 (p 2 = 10 3 . 3 3 . 13 2 . Extracting the square root, the rational part is 10 . 3 . 13 or 

 390, and the irrational part is \/10 . 3. Hence </> = 390 v^30. Eut the rational 

 part is of no use, and need not be calculated. It is sufficient to omit all numbers 

 of which even powers occur in the value of <£ 2 , and instead of any odd power of a 

 number to write the first power, or the number itself. Thus, instead of writing 



