+ Substituting in these functions x= -oo and w = too succes. sively, we obtain the following results, in respect to signs : X X RR { x = -00, + +, 3 variations. For c=+00, + + t +, 0 Hence, the given equation has 3 real roots. Since the signs in the given equation present two variations and one permanence, two of the roots must be positive, and the other negative, (446). To ascertain the situation of the positive roots, let us substitute in the functions, = 0, x=1, x = 2, etc., successively, noting the variations of signs in the results. 0, signs, + + 2 variations. x=1, + 2 1 variation, + 1 t, 1 5, + + + + 0 Since one variation is lost in passing from x= 1 to x = 2, and one also in passing from x=4 to x= = 5, one positive root must be situated between 1 and 2, and the other between 4 and 5. To ascertain the situation of the negative root, substitute x = = 0, x = -1, x = —2, etc.; the signs are as follows : 0, signs, + + 2 variations. -1, + 2 For + + + 2 + +, 2 -4, + + 3 Hence, the negative root is situated between -3 and -4. The initial figures of the several roots will be 1, 4, and —3. 2. Given the equation, X-2,9—7x* +10x+10 = 0, to find the number and situation of its real roots. In this example, we have X = x_2°—7.c*+10x+10, X, = 2.3.x*—7x+5, R= 1722–232—45, R = 152x—305, R2 = 524785. Let x = -00; we have + + ti 4 variations. x = +00; + + + + + 0 + 31 + X = Let 0 = + +, 2 -1; Hence, the roots are all real. And since the signs of the given equation give 2 variations and 2 permanences, two of the roots are: positive and two are negative. To find the situation of the roots, 0; we have + + + 2 variations. 1; + 2 3 + Hence, there are two roots situated between 2 and 3, one between 0 and –1, and one between —2 and —3. We wish now to find limits which will separate the two roots that lie between 2 and 3. Let us transform the given equation into another whose roots shall be less by 2. By (443), the operation will be as follows: 1 -2 -7 +10 +10(2 0 - 14 8 -3; +, 3 +, 4 66 + 6 = X'z 2.3 The transformed equation is therefore V=y+6y +5y9-10y+2 = 0. Now since the two positive roots of the original equation are found between 2 and 3, the two corresponding roots of the transformed equation must lie between 0 and 1. The situation of these roots may be found from V alone, by trials as follows: Substitute x = 0, .1, 2, 3, .4, .5, .6, -7, .8. The signs of V are +++ + Hence, one root of V lies between .2 and .3, and one between .7 and .8. Consequently the initial figures of the two positive roots in the original equation, are 2.2 and 2.7. NOTE.—If we had found the initial figures of the two positive roots of V to be the same, we should have proceeded to transform V, and make similar trials with the result. We are now prepared to find the roots of an equation to any degree of accuracy, by HORNER'S MÉTHOD OF APPROXIMATION. 463. In the year 1819, W. G. Horner Esq., an English mathematician, published a most elegant and concise method of approximating to the roots of a numerical equation of any degree. The process consists in a series of transformations, the roots of each successive equation being less than the roots of the preceding equation by the initial figures of the preceding roots. But in making the several transformations, the initial figures are obtained by trial division, as in square and cube root, and not by substitutions, as in the last article. 464. Let us take an equation in the general form, thus : X=2m + Arm-:+Bum-?to...+ Tc+U=0 (1) Let r represent the initial figure or figures of one of the real roots of this equation, as found by Sturm's Theorem, or otherwise. Now let the equation be transformed into another whose roots shall be less by r. erty; we shall have V = y + A'ym~+B'ym-'to...+T'y+U' = 0 In this equation y is supposed to represent a decimal, since r includes at least the entire part of the required root. Hence, the terms containing the higher powers of y are comparatively small; neglecting these, we have, approximately, U' T! Denote the first figure of this quotient by s; put y=s+z. Transforming eq. (2) into another whose roots shall be less by s, we have V' = "+A"zm-+B''xm-sto...+T":+U" = 0. Put x = Whence we have, as before, U" =t+7', T' where t is another figure of the required root. This process may be continued at pleasure, and we shall have, finally, c=rts+t+etc. Hence, to solve a numerical equation of any degree, we first find by Sturm's Theorem, or otherwise, the number of real roots, and also the first figure or figures of each. We may then approximate to the value of any root by the following RULE. I. Transform the given equation into another whose roots shall be less by the initial figure or figures of the required root. II. Divide the absolute term of the transformed equation by the penultimate coefficient, as a trial divisor, and write the first figure of the quotient as the next figure of the root sought. III. Transform the last equation into another whose roots shall be less than those of the previous equation by the figure last found; and thus continue till the root be obtained to the required degree of accuracy. NOTES. 1. The successive transformations required in obtaining any root may all be made in a single operation; and for the sake of perspicuity, the coefficients obtained in each transformation may be marked or numbered. 2. If a trial figure of the root, obtained by any division, reduces the absolute term X', and the penultimate coefficient X'1, to the same sign, this figure is not the true one, and must be diminished. 3. To obtain the negative roots, it will be most convenient to change the signs of the alternate terms of the given equation, and find the positive roots of the result; these, with their signs changed, will be the negative roots required. 4.-If the penultimate coefficient, T', should reduce to zero in the operation, the next figure of the root may be obtained by dividing the absolute term, X', by the coefficient which precedes T', and extracting the square root of the quotient. For, if T' vanishes, we have, in the transformed equation, EXAMPLES. 1. Given x* — 2x°—20x—40, to find the approximate value of x. By Sturm’s Theorem we find that this equation has only one real root, the initial figure being 6. We now obtain the decimal part, to 2 places, as follows: EXPLANATION.- We first transform the given equation into another whose roots are less by 6, using the method of Synthetic Division, explained in (443). The coefficients of the transformed equation are 16, 64 and —16, marked (1) in the operation. Divid. ing the absolute term —16, taken with the contrary sign, by the penultimate coefficient 64, we obtain .2, the next figure of the root. We next transform the equation whose coefficients are marked (1), into another whose roots are less by .2, the resulting coefficients being marked (2). Dividing 2.552 by 70.52, we obtain .03, the next figure of the root. The operation may thus be continued till the root is obtained to any required degree of accuracy. 2. Given 2*+x230x2—20x——20 = 0, to find one value of x. By Sturm's Theorem we find the initial figures of the two real |