262 SOLUTION OF RECURRING EQUATIONS. 484. The roots of a recurring equation of an even number of dimensions, exceeding a quadratic, may be found by the solution of an equation of half the number of di mensions. ...... Let x-px-1. which n is even; its roots are of the form a, - px + 1 = 0, be the given equation in 1 (Art. 455); or it may be conceived to be made up factors, a' 1 b, 7, &c. of quadratic (x − a) ( x − 1 ) ; (x − b ) ( x − ¦) ; &c. 1 that is, if a = a + B=b+ ; &c. of the quadratic factors, a x2 − a x + 1; x2- ßx+1; &c. Then, by multiplying these together, and equating the coefficients with those of the proposed equation, the values of a, B, &c. may be found. Moreover, for every value of each of the quantities a, ẞ, &c. there are two values of ; therefore the equation for determining the value of a will rise only to half as many dimensions as a rises to in the original equation. 485. If the recurring equation be of an odd number of dimensions, +1, or 1 is a root (Art. 458); and the equation may therefore be reduced to one of the same kind, of an even number of dimensions, by division. is obtained, which contains the other two roots, that is, the three roots of the equation a3 - 1 = 0, or the three cube roots of 1, are In the same manner, the roots of the equation 3 + 1 = 0 are found to be This also follows from Art. 445. Ex. 2. Let a - 1 = 0 Two roots of this equation are + 1, - 1; and by division, an equation which contains the other two roots, +√-1, and -√-1. Ex. 3. Let a + 1 = 0. that is, Assume (x2 − m x + 1) × (x2 − n x + 1) = x2 + 1; x1 − (m + n) x3 + (mn + 2) x2 − (m + n) x + 1 = x2 + 1, and by equating the coefficients, m + n = 0, and mn + 2 = 0 ; hence m = - m, and m2 + 2 = 0, or m2 = 2, and m = ± √√√2. Therefore the two quadratics, which contain the roots of the proposed biquadratic, are x2 a √2.x + 1 = 0, from the solution of which it appears that the roots are In the same manner may the roots of the equations a + 1 = 0, and a + 1 = 0, be found. [See Appendix II. Exs. 11, 12.] SOLUTION OF A CUBIC EQUATION BY 486. Let the equation be reduced to the form x3- qx + r = 0, (See Art. 447). where q and r may be positive or negative. Assume x = a + b, then the equation becomes or a3 + b3 + 3 ab × (a + b) − q × (a + b) + r = 0 ; and since we have two unknown quantities, a and b, and have made only one supposition respecting them, viz. that a+b=x, we are at liberty to make another; let 3ab equation becomes a + b + r = 0; q = O, then the also, since 3ab - 1 = 0, an equation of a quadratic form; and by completing the We may observe, that when the sign of √ one part of the expression, is positive, it is negative in the other, that is, 488. Let x3+6x-20=0; here q = x = √10+ √108 +10-108 = 2.732 -0.732 = 2. (Art. 316). COR. 1. Having obtained one value of x, the equa tion may be depressed to a quadratic, and the other roots found (Art. 432). 489. COR. 2. The possible values of a and b being discovered, the other roots are known without the solution of a quadratic. The values of the cube roots of a3 are (Art. 485). Hence it appears, that there are nine values of a+b, three only of which can answer the conditions of the equation, the others having been introduced by involution. These nine values are, In the operation we assume 3ab = q, that is, the product of the corresponding values of a and b is supposed to be possible. This consideration excludes the 2d, 3d, 4th, 5th, 7th, and 9th values of a + b, or ; therefore the three roots of the equation are the three cube roots of a3, the roots of the cubic are |