terms are fractional, we must begin by destroying the fractions; and this may always be done by substituting for x any quantity y, divided by a number which contains all the denominators of the fractions. For example, in the equation as we find here fractions which have for denominators 2, 3, and multiples of those numbers, assume x=2, and we shall 6 + =0, and multiplying by 64, 18 y1 — 3 y3 +12 y2 -162y+72=0. In order to determine whether this equation has any rational roots, it will be requisite to make trial of the divisors of 72; but as the roots may be either positive or negative, we must make two trials of each divisor. The following general rule will however enable us in most cases to dispense with this. Whenever the signs and succeed each other in regular + order, the equation has as many positive roots as there are changes in the signs; and as many times as the same sign recurs without the other intervening, so many negative roots belong to the equation. Now, the example given above contain four changes of the signs, and no succession; so that all the roots are positive, and we have only to make trial of the positive divisors. In the equation, x2+2x3-7x2-8x+12=0, we find two changes of signs, and also two successions; whence we conclude, with certainty, that this equation contains two positive, and as many negative roots, which must all be divisors of 12; viz. 1, 2, 3, 4, 6, 12. The first of these answers the terms of the equation, and we have therefore x=1 for one of the roots. Again, if we make x=2, we find this number also to satisfy the conditions of the equation; hence we have x=1, and x=2, for the two positive roots; and it remains, therefore, to find the two negative ones; which are found, upon trial, to be -3 and 4. Consequently the four roots of the equation are 1, 2, 3, and 4, and according to the rule given above, two of these roots are positive, and two negative. 451. Since, however, no root could be determined by this method, when the roots are all irrational, it became necessary to devise other expedients for finding the roots, whenever this case occurs, and two methods have been discovered for finding the roots of any equation of the fourth degree. Before these methods are explained, however, it will be proper to give the solution of some particular cases, which may often be applied with great advantage. 452. When the equation is such, that the coefficients of the terms succeed in the same manner, both in the direct and in the inverse order of the terms, as in the following: x2+mx3+nx2+mx+1=0, or in this other equation, which is more general: x2+ma x2+na2x2 + ma3x+a1=0, we may always consider such a formula as the product of two factors which are of the second degree, and are easily resolved. In fact, if we represent this last equation by the product, (x2+pax+a2) × (x2 + qax+a3)=C, in which it is required to determine p and q in such a manner that the above equation may be obtained, we shall find by expansion, x2+(p+q) ax3 +(pq+2)a2x2+(p+q) a3x+a1=0, hence we have, Now squaring the first of these equations we have p2+2pq +q2 m2; from which, if we subtract four times the second, or 4 pq 4n-8, we shall have, Having thus found p and q, and making each of the above factors0, we have x2+pax+a2=0, and x2+qax+a2=0. 1 The second factor gives x=- qa±-a.√q2-4, and these 2 2 are the four roots of the given equation. For example, let there be given the equation, x2-4x3-3x2-4x+1=0. Here a=1,m=-4, n=-3; consequently m2-4n+8=36, and/m"-4n+8=6. Consequently p=. 4 6 9= =-5, whence result the four roots: viz. 2 of which the two first are impossible, and the two last possible. 5±√21 2 453. The 453. The second case, to which this method of resolution is applicable, is the same as the first with regard to the coefficients, but different with respect to the signs, as in the equation, x2+max3+na2x2 — ma3x+a*=0, where the second and fourth terms have different signs. This equation may be represented by the product, (x2+pax—a2) (x2+qax—a2)=0, which becomes, by expansion, x2+(p+q) ax3+(pq−2) a2x2— (p+q)a3⁄4x+a1=0, from which, by the same operation as above, we shall have, and having thus found p and q, we shall obtain the four roots of the equation as before. 454. Let there be proposed, for example, the equation, x2-3×2x+3x8x+16=0. Here we have, a=2, m=-3, and n=0; so that Hence the two first roots are x=1±√5, and the two last CHAP. LVI. OF THE RULE OF BOMBELLI, FOR REDUCING THE RESOLU- 455. Ir has been already shown how equations of the third degree are resolved by the rule of Cardan, so that the principal object with regard to equations of the fourth degree is to reduce them to equations of the third; for it is impossible to resolve, generally, equations of the fourth degree, without the aid of those of the third; since when we have determined one of the roots, the others always depend upon an equation of the third degree. And we may hence conclude, that the resolution of equations of higher dimensions presupposes the resolution of all equations of lower degrees. Let there be given, then, the general equation of the fourth degree, x2+ax3+bx2+cx+d=0, in which the letters a, b, c, d, represent any possible numbers, and let us suppose that this equation is the same as (x2+ ¦ ¦ ax+p)2−(qx+r)3±0, in which it is required to de 1 2 termine the letters p, q, and r, in order to obtain the equation proposed. The last quantity by expansion becomes, Now, the first two terms are already the same here as in the H |