EXAMPLES. 1. It is required to exterminate the second term of the equation x2 + 3ax2 = b, or x3 + 3ax2 — b = 0. into another that shall want the second term. which is an equation where 22, or the second term, is wanting, as before. 10, be transformed into 3. Let the equation x3 another that shall want the second term. 4. Let y3 — 15y2 + 8ly Ans: 23-12z = 26. 243, be transformed into an equation that shall want the second term. Ans. x+6x=88. which shall want the third term, by assuming xy+e, and in the resulting equation, let 3e2 12e +9, or c2-4e30, in which the values of e are 1 and 3; then assume xy+3, or y+1, and the resulting equation will be y3+3y2—1 = 0, an equation wanting the third term. 4. The equation 6x2 1 11x2+6x-1=0, by assuming x= may be transformed into y36y2+11y-6=0; the roots of which are to be reciprocals of the former. 3 5. The equation 3x3 — 13x2+ 14x+16=0, by assuming x = may be transformed into y3-13y242y+1440, the roots of which are three times those of the former.-ED. ed into another that shall want the second term. 5x2+10x 4 = 0, be 6. Let the equation x2+8x3 transformed into another, that shall want the second term. Ans. y 29y2 + 94y — 92 92 = 0. 7. Let the equation x1 formed into another, that shall want the third term. 3x3 + 3x2 5x 2=0, be trans 8. Let the equation 3x3 Ans, ya + y3 — 4y 6=0. 2x + 1 = 0, be transformed into another, whose roots are the reciprocals of the former. Ans. y3 — 2y2 + 3 · = 0. 9. Let the equation +1x2 - 3x + 1 = 0, be transformed into another, in which the coefficient of the highest term shall be unity, and the remaining terms integers. Ans. y 3y+12y2 - 162 y + 72 = 0. — OF THE SOLUTION OF CUBIC EQUATIONS. RULE.-Take away the second term of the equation when necessary, as directed in the preceding rule. Then, if the numeral coefficients of the given equation, or of that arising from the reduction above-mentioned, be substituted for a and b in either of the following formulæ, the result will give one of the roots, as required.* * If, instead of the regular method of reducing a cubic equation of the general form x3+ax2 + bx + c = 0, to another, wanting the second term, as pointed out in the preceding article, there be put x=(y-a), we shall have, by substitution and reduction, y3 (9b3a2) y = 9ab27c-2a3; where, since the value of y can be determined by either of the formule given in this rule, the value of x will also be known, being x (ya). And if b = 0, or the original equation be of the following form, 23+ax2 + c = 0, the reduced equation will be y3-3azy 2a3-27c, where the value of y, being found as above, we shall have, as before, x= (y-a), which formulæ, it may be observed, are more convenient, in some cases, than those resulting from the preceding article; as the coefficients, thus obtained, are always integers.; whereas, by the former method, they are frequently fractions. Where, it is to be observed, that when the coefficient a, of the second term of the above equation, is negative, as 27' also, in the formula, will be negative; and if the absolute b term b be negative, in the formula will also be negative ; 2 It may likewise be remarked, that when the equation is of the form * This method of solving cubic equations is usually ascribed to Cardan, a celebrated Italian analyst of the 16th century; but the authors of it were Scipio Ferreus and Nicholas Tartalea, who discovered it about the same time, independently of each other, as is proved by Montucla, in his Histoire des Mathematiques, Vol. I. p. 568, and more at large in Hutton's Mathematical Dictionary, Art. Algebra. 'The rule above given, which is similar to that of Čardan, may be demonstrated as follows: b, Let the equation, whose root is required, be x3 + ax And assume y+z = x, and 3yx a. Then, by substituting these values in the given equation, we shall have y3+3y2z+3yz 2 +z3 + a × (y + z) = y3+z3+3yz × (y + z)+a× (y + z) = y3 +23 a × (y + z ) + a×(y+z)=b, or y3+23=b. And if, from the square of this last equation, there be taken 4 times the cube of the equation yz ta, we shall have y6-2y323+z6 = b2+27 a3, or y3—23 = √ (b2+21703). 4 4 But the sum of this equation and yз+23=b, is 2y3=b+√ (b2 + 2 a3), and their difference is 223 — b 6 — √ (63 + 27 a3); whence y √(63+37 =3/ [bb + √ (163 +27 a3)], and z =3/ [b — √ (162 +27 a3)]. 3 From which it appears, that y+z, or its equal x, is √ [16+√(162+27α3)] + V [18-√(b2+7a3)], which is the theorem; tion of it cannot be obtained by the above rule; as the question, in this instance, falls under what is usually called the Irreducible Case of cubic equations.* 1. Given 2x2 Here x3 EXAMPLES. 12x2 + 36x = 44, to find the value of x. 6x2 + 18x: 22, by dividing by 2. And, in order to exterminate the second term, And consequently, by substituting 6 for a and 2 for b, in the first formula, we shall have 18z+36 22, or 23 + 6z 23 * It may here be farther observed as a remarkable circumstance in the history of this science, that the solution of the Irreducible Case above-mentioned, except by means of a table of sines, or by infinite series, has hitherto baffled the united efforts of the most celebrated Mathematicians in Europe; although it is well known that all the three roots of the equation are, in this case, real; whereas in those that are resolvable by the above formula, only one of the roots is real; so that, in fact, the rule is only applicable to such cubics as have two equal, or two impossible roots. The reason why the assumptions, made in the note to the former part of this article with respect to the solution of the equation x3 ax=b, a are found to fail in the case in question (and it does not appear that any other can be adopted) is, that the two auxiliary equations, 3yz: and y3+23=b,“which in this case become 3yz=a, and y3+23—b, аз 27 or y3z3 and y3+z3=b, cannot take place together; being inconsistent with each other. For the greatest product that can be formed of the two quantities y3+23 is when they are all equal to each other; or since y3+ 23 = 6, when each of these b; in which case their product is 162. But, as above shown, y3z3 аз аз by the question; therefore, when 27 b2 4. the two conditions are incompatible with each other; and conequently the solution of the problem, upon that supposition, can only be obtained by imaginary quantities. 12 3 3 V[1 + v(1+8)] + 33⁄4√ [1 − √ (1 + 8)] = √(1 + 3⁄4/ 9) + 3 3/(1 − √9) = (1 + 3) + 3 (1 − 3) = 3/4-3/2, Here a being equal to — 6, and b equal to 12, we shall Therefore x = 3.1392, the answer. 3. Given x3 2x: 4, to find the value of x. Here a being formula, - 2, and b 4, we shall have, by the 3 1.5773 1.9999, or 2; Therefore x = -2, the answer. // (−2+1.9245) (2 +1.9245)=3(-.0755 3/3.9245 === .41226 * 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 a±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. 3 Or if, as in the above example, the roots, or numeral values of 3/ 10 (−2+ √3), and — √(2+3), be determined according to 9 10 the rule laid down in Surds, Case 12, the result will be found equal to -2 as it ought. |