II, DERIVÉD EQUATICNS AND EQUAL Roots. Definitions. 204. A derived polynomial, is one that may be derived from a given polynomial, by multiplying each term by the exponent of the leading letter in that term, and then diminishing the exponent of the leading letter, by 1, in each result; the derived polynomial is also called the derivative; and that from which it is derived, is called the primitive. Thus, 23 + 2.32 + 3x + 1, being a given primitive polynomial, its derivative is, 3x2 + 4x + 3. A derived equation, or a derivative equation, is one whose members are derivatives of the two members of a given equation. Thus, 3.x2 + 2x + 5 = 0, being a primitive equation, its derivative is, 6x + 2 = 0. EXAMPLES. Find the derivatives of the following equations : 1. 24 + 3x2 + 2 = 0. Ans. 4x3 + 6x = 0. 2. 24 + 6x3 + 20x2 + 10 = 0. Ans. 4x3 + 18x2 + 40x = 0. The derivative of the equation, 2n + pan-1 + qxn-? + ... + 8x2 + tw + U = 0, . (1) is, nxn-1 + (n-1)pxn—+ (n--2) qan–3+ + 25x + t = 0. . (12). Equal Roots. 205. We have seen that the first member of equation (1) is composed of as many binomial factors of the form, X a, X – b, &c., as there are roots, each of which corresponds to a root of the equation. When the equation has two roots equal to a, there will be two factors equal to X a, that is, the first member will be divisible by (x – a); when there are three roots equal to a, the first member will be divisible by (x – a), and so on. To deduce a general rule for determining whether an equation has equal roots, and for freeing it of them, let us resume the general equation, 2012 + pan-1 + qan-3 + ... + sä? + tx + u = 0; . (1) substituting y + p for x, we have, 0; (y + r)" + p(y + r)n-1 + 9 (y + r)n-2 +...+$(y + r)? +t(y + r) +u = developing the different powers of y + r, by the binomial formula, and arranging the result according to the ascending powers of y, we have, in this equation, the values of y are equal to the values of x, in equation (1), each diminished by r, since, by hypothesis, we have, X = y + r; whence, y = x Now, r is entirely arbitrary; we may therefore assign a value to it at pleasure. If we supposer to be a root of equation (1), the coefficient of yo, in equation (a), will be equal to 0, because that coefficient is what the first member of (1) becomes, when r is substituted for x (Art. 193); making this coefficient equal to 0, and dividing each of the remaining terms by y, we have, Now, then – 1 roots of equation (6), are equal to the different values of x _ r, in equation (1), that is, to the results obtained by subtracting the root r of equation (1), from each of the other roots in succession. If, therefore, equation (1) has two roots equal to r, one of these differences will be equal to 0, that is, one of the roots of equation (6), will also be equal to 0. But, when x-r is equal to 0, that is, when y is equal to 0, all of the terms of equation (6), except the first, reduce to 0, consequently that term must be separately equal to 0; or, npn-1 + (n 1)ppn? + (n − 2)q-3 + (c). Comparing this equation with equation (12), (Art. 204), we see that it is what the derivative of equation (1) becomes, when for a we substitute r. Hence, we conclude that when equation (1) has two roots equal to 1, its derivative has 1 root equal to r; and conversely, when the derivative has 1 root equal to r, the primitive has 2 roots equal to r. By a similar course of reasoning, it may be shown that, when equation (1) has 3, 4, 5, &c., roots, each equal to r, its derivative has 2, 3, 4, &c., roots, each equal to r; and conversely, when the derivative has 2, 3, 4, &c., roots, each equal to r, the primitive has 3, 4, 5, &c., roots, each equal to r; that is, when the first member of the given equation is divisible by (x − p), (x — )}, &c., the first member of its derivative will be divisible by 2 – 1, (2 — )?, &c., and the reverse; hence, we have the following rule for determining whether an equation has any equal roots; and for freeing the equation from them if it has any: RULE. Find the derivative equation; then apply the rule for the greatest common divisor to the first members of the primitive and derivative equations; if no common divisor is found, the equation has no equal roots; if a common divisor is found, divide both members of the given equation by it, and the resulting equation will have no equal roots. The operation of finding the values of equal roots, consists in placing the greatest common divisor found equal to 0, and solving the resulting equation. When this can be done, the equal roots may all be found. For each single root of the resulting equation, there will be two equal roots in the primitive equation ; for each pair of equal roots in the resulting equation, there will be three equal roots in the given equation; and so on, as indicated in the preceding discussion. The equation found, by placing the greatest common divi. sor equal to 0, is to be treated, in all respects, like an original equation; and of course, the process for equal roots may be applied to it; and so on, indefinitely. EXAMPLES. 1. Eliminate the equal roots from the equation, 205 — 2723 + 2222 + 192x + 288 = 0. The derivative equation is, 5.24 81x2 + 44x + 192 = 0. The greatest common divisor of the first members of the given equation and its derivative, is, (x – 3) (20 + 4); dividing both members of the given equation by this, we have the equation, 203 — X2 – 14x + 24 = 0, which has no equal roots. If we place the common divisor equal to 0, we have, (x − 3)(x + 4) = 0. .. X = 3, and 2 = - 4. Hence, the given equation has two roots equal to 3, and two roots equal to – 4. Dividing both members of the given equation, by (2x – 3)? (x + 4)?, there results, The given equation is completely solved, and, in like manner, many other equations may be treated. 2. Eliminate the equal roots from the equation, 5x3 + 9x? — 7x + 2 = 0. The derivative equation is, 4X3 – 15x2 + 18x – 1 = 0; |