Commensurable Roots of any Equation. m 32, 22, 20; 30; -25; =-40; р 5, 4, 2, 1, 1, 2, 4, 5; m = 4, 2, 8, -28, 52, m" 30, 6, -18, 18, m'' = 4, — 13, — 1, -28, m1 = 2, — 13, 1, 14, 5; and, therefore, 2, — 1, and -5 are roots of the given equation, and its first member, divided by the factor (x-2)(x + 1)(x+5) = 23+42?–71—10, gives the quotient 22 - 4x + 4; and, therefore, the remaining roots are those of the equation 22 - 4x +4= 0, which are equal to each other, and each is X = 2. 2. Find the commensurable roots of the equation 38 - 3x7-10 26 —2z4+673 +21.22—31-10= 0. Ans. 5, 1,-1, and — 2. 3. Find all the roots of the equation 24 + 23 - 24 x2 + 43 21 = 0 which has commensurable roots. Ans. 1, 3, -53. 4. Find all the roots of the equation 23 — 6x2 + 19 x — 44 = 0 which has a commensurable root. Ans. 4, and 1 EV - 10. 5. Find all the roots of the equation 24 — 10 23 +35 % - 50 + 24 = 0 which has coinmensurable roots. Ans. 1, 2, 3, 4. Commensurable Roots of any Equation. 6. Find all the roots of the equation 25 - 32* - 823 +242 — 91 +27= 0 which has commensurable and equal roots. Ans. 3, -3, and EV - 1. 7. Find all the roots of the equation 26 - 234 - 48 23 +95 +2 +400x+375=0 which has commensurable and also equal roots. Ans. 3, 5, and — 2+-1. 313. Problem. To find the commensurable roots of an equation. Solution. Reduce the equation to the form AI" + Bx*-1 + &c. ...+La+M=0, in which A, B, &c., are all integers, either positive or negative. Substitute for x the value + +-&c.... + +M=0; A' and the equation becomes Byn-1 Cy-2 Ly + An-1 An-1 Anh A which, multiplied by An-1, is ya+Byn-1 +A Cyn-2+&c.... + An-2 Ly+An-1M=0. The commensurable roots of this equation may be found, as in the preceding article, and being divided by A, will give the commensurable roots of the required equation. 314. Scholium. The substitution of y A Commensurable Roots of any Equation. is not always the one which leads to the most simple result. But when A has two or more equal factors, it is often the case that the substitution y A leads to an equation of the desired form, A' being the product of the prime factors of A, and each factor need scarcely ever be repeated more than once. 315. EXAMPLES. 1. Find the commensurable roots of the equation 64 24 — 328 23 + 574 22. 393 27-90=0. Solution. We have, in this case, A= 64 = 26; hence we may take A' equal to some power of 2; and it is easily seen that the third power will do, so that we may make =fy. Hence the given equation becomes y4 — 41 y3 + 574 y2 - 3144 y + 5760 = 0. The commensurable roots of which are found, as in art. 311, to be y 4, 6, 15, and 16; so that the roots of the given equation are 1, 1, 13, and 2. 2. Find the commensurable roots of the equation 8.23 +34z? — 79+30=0. Ans. j, k, and — 6. X = Commensurable Roots of any Equation. 3. Find the commensurable roots of the equation 24 23 26 22 + 9x—1= 0. Ans. 1, 3, and 4. Find the commensurable roots of the equation 3 23 — 14 x2 + 21 x - 10 = 0. Ans. 1, $, and 2. 5. Find the commensurable roots of the equation 8 24 - 38 23 +49- 22x +3=0. Ans. 1, 1, 1, and 3. 6. Find all the roots of the equation 6.23 +7x2 + 39 x + 63 = 0 which has a commensurable root. Ans. - *, and tĖt-251. 7. Find the commensurable roots of the equation 926 +30 25 + 22 24 +10 23 +17x2—20x+4= 0. Ans. $ and -2. Value of Continued Fractions. CHAPTER IX. CONTINUED FRACTIONS. 316. A continued fraction is one whose numerator is unity, and its denominator an integer increased by a fraction, whose numerator is likewise unity, and which may be a continued fraction. Thus, 1 1 and 1 a + 1 6+ 1 otat &c. are continued fractions. 317. Problem. To find the value of a continued fraction which is composed of a finite number of fractions. Solution. Let the given fraction be 1 1 at 1 1 bt cta Beginning with the last fraction, we have successively 1 cd+1 d d |