V[1 + v(1+8)] + V[1 − v (1+8)] = √(1+9) + /(1-9)=√(1 + 3) + 3/(1-3)=4-3/2, Therefore r = 2 + 2 = 3/4-3/2+2=2+1.587401 1.259921= 2.32748, the answer. 9 9 (-2+1.9245) (2 + 1.9245)/(-.0755 — V 3.9245 .412261.5773 : = 1.9999, or - 2; Therefore x = - - 2, the answer. Note.-When one of the roots of a cubic equation has been found, by the common formula as above, or in any other way, the other two roots may be determined as follows: Let the known root be denoted by r, and put all the terms of the equation, when brought to the lefthand side, = 0; then, if the equation, so formed, be divided by x±r according as r is positive or negative, there will arise a quadratic equation, * When the root of the given equation is a whole number, this method only determines it by an approximation of 9s, in the decimal part, which sufficiently indicates the entire integer; but in most instances of this kind, its value may be readily found, by a few trials, from the equation itself. 10 10 Or if, as in the above example, the roots, or numeral values of 3/ (−2+—√3), and −√(2+3), be determined according to the rule laid down in Surds, Case 12, the result will be found equal to - 2 as it ought. 9 the roots of which will be the other two roots of the given cubic equation. 4. Given x3- 15x = 4, to find the three roots or values of x. Here x is readily found, by a few trials, to be equal to 4, and therefore 1. Given + 3x2. tion, or the value of x. 2. Given 3 + x2: the value of x. 3. Given x3 the value of x. 4. Given 3 the value of x. = 6x=8, to find the root of the equaAns. x= = 2. 500, to find the root of the equation, or Ans. x 7.616789. 3x2=5, to find the root of the equation, or Ans. x= : 3.425101. 6x=6, to find the root of the equation, or Ans. 4+ V2. 5. Given 3+9x= 6, to find the root of the equation, or the value of x. 6. Given 3 + 2x2 tion, or the value of x. 23x= Ans. 9 3/3. 70, to find the root of thee quaAns. x5.134599. 350, to find the root of the Ans. = 14.954068. 4, to find the three roots of the equation, or the three values of x. 7. Given 23 17x2+54x= equation, or the value of x. 8. Given 3 6x 9. Given a3 Ans. ✓ 3. - 2, 1+v 3, and 1 12, to find the three roots of the equation, or the three values of x. Ans. 3,1+v5, and 1 — √5. OF THE SOLUTION OF CUBIC EQUATIONS, BY CONVERGING SERIES. THIS method, which, in some cases, will be found more convenient in practice than the former, consists in substituting the numeral parts of the given equation, in the place of the literal, in one of the following general formulæ, according to which it may be found to belong, and then collecting as many terms of the series as are sufficient for determining the value of the unknown quantity, to the degree of exactness required.* 1. x2 + ax = - b.t The method laid down in this article, of solving cubic equations by means of series, was first given by NICOLE, in the Memoirs of the Academy of Sciences, an. 1738, p. 99; and afterwards, at greater length, by CLAIRAUT in his Elemens d'Algebre. With respect to the determination of the roots of cubic equations by means of series, let there be given, as above, the equation x3+ax, b, where the root by transposing the terms of each of the two branches of the common formula, is x = √ √/ { √ (462 + 11, a3) + 16 } — 3/ { √ (462 +27 α3) — 16}; or by putting, for the sake of greater simplicity, (62 +27 a3) =s, and reducing the expression, x = s3 Hence, extracting the roots of the righthand member of this equation by the binomial theorem, there will arise 3.6.9 (3) 3 2.5.8 b (2) ' -, &c. 2.5.8 3.6.9.12 And consequently, if the latter of these two series be taken from the former, the result, by making the first term of the remainder a multiplier, will give 26 6s3 Whence, also, by substitution, we have the above formula In which case, as well as in all the following ones, A, B, C, &c., denote the terms immediately preceding those in which they are first found. a3, or 276a > 4a3. x=±23/ b 1 2 b, where 162 is supposed to be greater than 2,2762-4a3, (2762 3.6 2762 2762 4a3 ·). 3 &c. * Or, 4a3 5.8,2762 2762 2 2762 17.20 ) c 15.18 2762 21.24 2762 9.12 2762 -); &c. D In which case the upper sign must be taken when b is *The root, as found by the common formula, when properly reduced, is b Whence, extracting the roots of the righthand member of this equation, there will arise 3/(1+s)=1+ts--; And, consequently, by adding the two series together, and taking the 1st term of the result as a multiplier, we shall have x=2 / 2 positive, and the under sign when it is negative; and the same for the first root in the two following cases :— ±b, ax= where 162 is supposed to be less than 2 a', or 276a < 4a3. *This expression is obtained from the last series, by barely changing the signs of the numerator and denominator in each of its terms; which does not alter their value. Hence, in order to determine the other two roots of the equation, let that above found, or its equivalent expression, √3b+√(462 — 227 + 3/36 — √ (1b2 — 2, a3)} = ±r. Then, according to the formula that has been before given for these roots in the former part of the present article, we shall have x = ± putting 2 r 2 2 (b2 — 27 a3) = s, and reducing the expression, x === Or, {3⁄4/(1+s)—V/(1—s)}. Whence, extracting the cube roots of the righthand member of this equation, there will arise 3.6 2.5 $3 3.6.9 And, consequently, by taking the latter of these series from the former, and making the first term of the remainder a multiplier, we shall have (443-3763), &c., and also V » × √ −3 = { '4 2.5.8.11.14.17 ㄓ 2 3 2 b 2762-4a3 s2= 2762 -3 31⁄2 2762 if these values be substituted for their equals, in the last series, the result will give the above expressions for the two remaining roots of the equation. V-3 2762 (2762-4a3 962 |