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 is readily found, by a few trials, to be equal to 4, and therefore 6x 8, to find the root of the equaAns. x 2. tion, or the value of x. 2. Given x3 + x2 - 500, to find the root of the equation, or the value of x. 3. Given x3 the value of x. 6x Ans. x7.616789. 3x2 = 5, to find the root of the equation, or Ans. x= 3.425101. 4. Given 3 6, to find the root of the equation, or the value of x. Ans. 4+ 3⁄4//2. 5. Given æ3 + 9x = 6, to find the root of the equation, or the value of x. Ans. 3/9 3/3. 6. Given + 2x2 23x= 70, to find the root of thee quation, or the value of x. Ans. x5.134599. +54x=350, to find the root of the Ans. 14.954068. 8. Given a3 6x4, to find the three roots of the equa 7. Given a3- 17x2+54x= equation, or the value of x. tion, or the three values of x. Ans. 2, 1+ √ 3, and 1 — √ 3. 9. Given x3 5x2+2x= 12, to find the three roots of the equation, or the three values of x. Ans. 3,1+v5, and 1 — v5. 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. * 26 1. x23 + ax = b.† 1 +230 (27 + 6.9 \2762 + 4a3. 6.9.12.15 * 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 xз+ax Eb, where the root by transposing the terms of each of the two branches of the common formula, is V { √ (+b2 + 317 a3) + 10 } — 2/ { √ (162 + 27 a3)— 36 } ; or by putting, for the sake of greater simplicity, v (3b2-+-27 a3) = s, and re b ducing the expression, x = 5* {V(+2)—v(1−2)}. s Hence, extracting the roots of the righthand inember of this equation 3 by the binomial theorem, there will arise /(1+1)=1+(82) 2 — 3.6 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 h 2b 2*762 27b2+4a3' Whence, also, by substitution, we have the above formula 2.5.8.11.14.17 2762 3 6.9.12.15.18.21 (2762 + 4a3) 3 +, &c. } Or 3/ [2(2762 + 4a3)] {1+ 2763 8.11 +( 6.9\2762 + 4a3 12.15 14.17 2762 (2702 4a2) 12.21 \2762+4a3 (2762 + 4a3) D +, &c. 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. 2. x3 ax ±b, where 162 is supposed to be greater than 24a3, or 2762 > 4a3. In which case the upper sign must be taken when b is * The root, as found by the common formula, when properly reduced, is =s, we shall have = {v/(1+s)+v (1−s)}. 2 Whence, extracting the roots of the righthand member of this equation, 2 $3 3.6.9 3.6.9.12 &c. 3/(1—s)=1— }s⋅ $2 $3 S4, &C. there will arise 3/(1+s)=1+}s 3.6 3.6.9 3.6.9.12 And, consequently, by adding the two series together, and taking the 1st term of the result as a multiplier, we shall have x positive, and the under sign when it is negative; and the same for the first root in the two following cases :— where 162 is supposed to be less than 1 a3, or 27b2 < 4a3. 27 2.5.8 4a3-27b2 ( 3.6.9.12 2762 2 ( -) 3.6 -2762 -)2 * 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, v{+√(4b2—27 a3) } 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 = 干 ㄓ 3 2 2 2 putting (162-27 a3) =s, and reducing the expression, x = I 2 {3⁄4/(1+s)—V(1—s)}. Whence, extracting the cube roots of the righthand member of this equation, there will arise 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 H 2.5 2.5.8.11 6.9 6.9.12.15 6.9.12.15.18.21 2.5.8.11.14.17 •s+, &c. }• (403—2763)3, &c., and also V×_v_−3=VI 4a3-2762 962 √(4a3—27b2) 93/ 2b2 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. 21.24 ( -2763) 9.12 2762 12763) D+, &c. 2762 which series answers to the irreducible case, and must be used when 2a3 is less than 2762. And if the root thus found be putr, the other two roots may be expressed as follows: γ √(4a3 — 272) 2 { 2.5/4a3 2762 2.5.8.11 1 + 6.9 2762 6.9.12.15 2762 2.5/4a3 2762 8.11 A + 6.9 2762 12.15 20.23/4a3 2762 B c+ 24.27 2762 '4a3 — 27b2. 14.17/4a3- 2762 18.21 2762 Where - r, or +ar, must be taken according as b is positive or negative; and the double signs must be considered as in one case, and in the other, as usual. where 162 is still supposed to be less than a3, or 2763 <4a3. * By transposing the terms of the common formula, as in the first case, we shall have x = 1 3/ { √ (3b2 — 2+, a3) + 36 } − 3) { √ (3b2— 27 as) 7 -16. Or, by putting, for the sake of simplicity as before, V. (462 —,, a3)=s, and reducing the equation = /s{/ (1 + 2/1) - V Whence, extracting the roots of the righthand number, as in the former instances, ~ (1+2)=1++(2) — 2. (2)2 + |
