The irrational part will be the algebraic sum of the terms having the odd powers of Vb for factors. But since (10)' = bvb, (vb)' = l’//b, etc., the different terms of this part can be represented by a single term of the form Nb, N being the algebraic bum of the coefficients of vb. Hence equation (2), under the supposition that a+vb is a root of equation (1), becomes M + NVb= 0; (3) which can be true only when we have separately M=0, N= 0; (272, 4). In reducing equation (2) to equation (3), the upper signs in the expansions of the terms of equation (2) were used. If the lower signs in the equation and the expansions of its terms be used,—which is equivalent to supposing a-n to be a root of equation (1),—the reduced equation will be M-Nb= 0. in which M and N are evidently the same as in equation (3). Hence if equation (1) has a root, aty/b, it has also the root am Now let us suppose that u+V -b is a root of eq. (1); then since the even powers of V -b are real and the odd powers imaginary, the developed first member of eq. (2) will be composed of two parts, the one real and the other imaginary. Represent the real part by M'. The imaginary part is the algebraic sum of the terms having the odd powers of V – for factors. But since (v —b) = V 6° (46) =bV6, (1-0)=VW(-5) =lVb, etc., the different terms of this part can be reduced to a single term of the form N'v=b. Hence, under the supposition that atv is a root of equation (1), equation (2) becomes M'+N'V-6 = 0, (5) which requires that we have separately M'=0, N'=0; (267). By using the lower signs of the terms and their expansions in equation (2),—which supposes a- -V6 to be a root of eq. (1),—we find M'-NV-b=0; (6) and by a simple inspection of the expanded terms of equation (2), we see that M' and N' in equations (5) and (6) are the same. Whence we conclude that if equation (1) have a root, a tv.in it has also the root, a-V-6. RULE OF DES CARTES. 3d. 4th. 6th, 9th, 1st. 2d. 446. An equation can not have a greater number of positive roots than there are variations in the signs of its terms, nor a greater number of negative roots than there are permanences of signs. NOTE. - A variation is a change of sign in passing from one term to another; a permanence occurs when two successive terms have the same sign. It is obvious that the number of variations and permanences taken together must be equal to the number of terms, less 1. Let the signs of the terms of an equation be + + + + + the second, sixth, and eighth terms giving permanences, and the other terms variations. To introduce a new positive root into the equation, we must multiply the equation by some binomial factor in the form of x-a; and the signs of the partial and final produets will be as follows: 7th, 8th. + + + + + + + + + + + + + Now we observe, in the final result, that the 2d, 6th, and 8th terms, or the terms which give permanences in the proposed equation, are ambiguous; consequently, let these ambiguous signs be taken as they may, the number of permanences has not been increased. But the number of terms has been increased by 1; hence, the number of variations has been increased by 1, at least. Again, to introduce a new negative root into the proposed equation, we must multiply by some binomial factor in the form of uta: and the partial and final products will be as follows: 8th. 9th, + + + + + + + + + + + + + + + + + Here, in the final result, the 3d, 4th, 5th, 7th and 9th terms, oi the terms which give variations in the proposed equation, are am. + 1st. 2d. 3d. 4th. 5th, 6th, 7th, It biguous; consequently, the number of variations has not been increased. But the number of terms has been increased by 1; hence, the number of permanences has been increased by 1, at least. Thus we have shown that the introduction of each positive root must give at least one additional variation, and the introduction of cach negative root must give at least one additional permanence. Hence the whole number of positive roots can not exceed the numker of variations, and the whole number of negative roots can not exceed the number of permanences; the proposition is therefore proved. 447. Although the introduction of a positive root will always give an additional variation of signs, it is not true that a variation of signs in the terms of an equation necessarily implies the presence of a real positive root. Thus, the equation, -*_7x+15 = 0 has 2 variations of signs, and 1 permanence. But its roots are 2+1=1, 2–1-1, and -3, no one being positive and real. But when the roots are all real, the number of positive roots is equal to the number of variations, and the number of negative roots is equal to the number of permanences. CARDAN'S RULE FOR CUBIC EQUATIONS. 448. It has been shown, (437), that any equation can be transformed into another which shall be deficient of its second term That is, every cubic equation can be reduced to the form of x+3px = 29; (1) and the solution of this equation must involve the general solution of cubics. We make 3p the coefficient of x, and 2q the absolute term, in order to avoid fractions in the following investigations: Assume x=v-ty; then eq. (1) becomes (v+y)+3p(u+y)= 29. (2) Expanding and reducing, we have vi+y* +3(vy+p)(v+y)= 29. (3) Now as the division of x into two parts is entirely arbitrary, we are permitted to assume that vy+p = 0; (4) whence, from eq. (3), vity' = 29. (5) If we obtain the value of y from (4), and substitute it in (5), we shall have, after reducing, 0°—2qv =p'; (6) whence, vi = 9+V q*+p'. () Substituting this value of v' in (5), we have yo = qFV q*+pa. (8) But by hypothesis x = v+y; hence, taking the sum of the cube roots of (7) and (8), we have (q+V2+po)+(-V?+pe)(A) which is Cardan's formula for cubic equations. 449. When p is negative, in the given equation, and its cube numerically greater than q*, the expression V q*+p' becomes imaginary; this is called the Irreducible Case. We must not conclude, however, that in this case the roots of the equation are imaginary; for, admitting the expression V q*+på to be imaginary, it can be represented by aV -1; whence the value of , in formula (A) be comes (1) =(q+av=1)*+(q-av=1); ==qd(14v=) *49*(=v=) *; or, (2) or, (3) -1 --1 9 Now by actually expanding the two parts in the second member of (3), and adding the results, the terms containing V-1 will cancel, and the final result will be real. Hence, in the irreducible case all the roots of the equation are real; formula (A) is therefore practically applicable only when two of the roots are imaginary. In this case the real root can be found directly by the formula; the equation may then be depressed, by division, to a quadratic, which will give the two imaginary roots. EXAMPLES. 2 = 1. Find the roots of the equation, X* — 7x* †14x—20 = 0. To transform this equation into another deficient of its 2d term, according to (437), put x=y+}; and we shall have for the transformed equation, y-jy=3144 To apply the formula to this equation, we have 3p =-, or p= -3; 2q = 44, or q = 72. =+= =$+; 5, the real root. Dividing the given equation by 2–5, we obtain for the depressed equation 3* —2+4= 0; whence, w = 1+V3. Hence the three roots are 5, 1+1=3, and 1-13. 2. Given x: +6x = 88, to find the values of w. To apply the formula, we have 3p = 6, or p = 2; or q = 44. whence, Va+p=V1944+8 = +44.090815+. And we have x'= (44+44.090815)#+(44–44.090815)#; or, - 4.4495—.4495 4, the real root. The depressed equation will be ac* +4x—22 = 0; whence x = -2+31-2; and the three roots are 4, —2+31=2, and —2–31–2. 3. Given XS_6x 5.6, to find one value of x. This example presents the irreducible case; the solution, by the method of series, is as follows: = |