CUBIC EQUATIONS. 581 



It makes no difference whether the given cubic wants the second term 

 or possesses it. In the latter case, it must be taken away by the usual rule, 



which will change the root from a + Vb to (a — ^J + Vb (p being the coefficient 



of the second term.) If p is non-divisible by 3, this will cause the new equa- 

 tion to have fractional coefficients. But these cause no difficulty. See examples 

 8 and 9. 



I have sufficiently shown how roots of the form a + \/b can be found, pro- 

 vided that Vb is a surd in its lowest terms. I will now proceed to consider roots 

 of the form a + kVb, where, for simplicity, I will suppose a, b, k integers, \/b a 

 surd in its lowest terms, and that the equation wants its second term. 



Theorem. — "k is always a factor of |\" 



For, since (p is the product of the differences of the roots, if the roots are 

 a + k Vb, a — k «/b, — 2 a, the differences taken in order will be 2 k Vb, 3a — Wb, 



— 3a — k */b, the product of which is 2 k */b (k 2 b — 9 a 2 ) — $, whence $ = Wb 



(k 2 b — 9 a 2 ), of which it is evident that k is a factor. It is also evident, that by 

 computing the value of <p, Vb becomes known, although a and k continue unknown. 



But, since k is a factor of ^, it may sometimes be easily discovered. This will 



best be explained by a few examples. It must be remembered, that since the 

 roots x, y, z, are a^= k Vb, and — 2 a, the coefficient q — xy + xz+ yz = xy 



— {x + yf = a 2 — k 2 b — 4 a 2 = — 3 a 2 — k% whence 3 a 2 + k 2 b = — q. 



Example 1. Let x % — 245 x — 482 = 0. 



Here <p 2 = — (4# 3 + 27 r 2 ) = 52551752, which, being decomposed in the usual 



way, we find J = 11233 \/2, from whence we learn that b = 2, and that the 



roots are therefore of the form a d= k \/% Since then k is a factor of ^, it must 



either equal 11 or 233. But it must also satisfy the general equation 3 a 2 + bk 2 

 = — q, which in this instance is 3 a 2 + 2 k 2 = 245 ; and since the number 233 

 is evidently much too large, the true solution must be k = 11. Hence 3 a 2 = 245 



— 2 k 2 = 245 — 242 = 3, whence a = 1. Therefore, the three roots are 1 + ll\/2 

 1 — 1 W% and — 2. 



Before going further, I wish to make some general remarks. All cubic equa- 

 tions, whose coefficients are whole numbers, may, I believe, be divided into three 

 classes. 



(1.) All the roots integers. 



(2.) Only one root an integer. 



(3.) No integer roots. 



It is of the second class only that I am now treating. Of course, if an equation 



VOL. XXIV. PART III. 7 S 



