We have or, or, 2.8 Put p=-2, q= 2.8; hence, (1+?v=1)+(1–1V=1) b = ļV-1; then l'=-d's, b= 15X7's. Also, (1+?V=1)+ = (1+0)*; (1-?V=1)+=(1–0). By the binomial theorem (1+0)* =1+34 =1+34-36 (+3-6-9-3-6-9.126*+ 56*+... (1–0) =1-36-3.662–3-6-90?– 3-6-9-12% = 2-2(36* )--(36912) ) = 2+.004535--000034= 2.004569. Hence, we have = 2.004569; x = (2.004569) 8/2.8 = 2.82535, Ans. 2 2-5 2.5.8 1 2 2.5 2.5.8 Sum 3 2.8 4. Given -6x-6 = 0, to find one value of x. Ans. x = 32+34= 2.8473+. 5. Given x +9.0—6 = 0, to find one value of x. Ans. x = 39+03 = .63783+. 6. Given x +6x9--13x+24 = 0, to find the values of x. Ans. r = -8, 1+1=2, or 1–V2. NUMERICAL EQUATIONS OF HIGHER DEGREES, 405 SECTION IX. SOLUTION OF NUMERICAL EQUATIONS OF HIGHER DEGREES. LIMITS OF REAL ROOTS. 450. All positive roots of an equation are comprised between 0 and + oo, and all negative roots between 0 and 00. But in the solution of numerical equations of higher degrees, it is necessary to be able at once to assign much narrower limits. As preliminary to this, we will first show how an equation is affected by substituting for the unknown quantity numbers greater or less than the roots, and numbers between which the roots are comprised. 451. If an equation, in its general form, be regarded as the product of the binomial factors formed by annexing the roots, with their opposite signs, to x, we observe that the sign of this product can not be affected by the imaginary roots. For, according to (445), if an equation have one root in the form of a+V –b, it will have another in the form of a-v-. But we have (x—a-1-6) (x—a+126) = (r—a)'+b, a result which is in all cases positive. 452. Let a, b, c, d, etc., be the real roots of an equation, ar. ranged in the order of their algebraic values; then the equation may be represented as follows: (a—a)(x-1)(x—c)(x—d).... If we substitute h for x, the first member will become (h—a)(h—b)(h—c)(h—d).... Now if h be less than the least root, a, every factor will be neg. ative; and the whole product will be positive or negative, according as the number of factors is even or odd. But the number of factors is equal to the degree of the equation, (427); hence, 1.-If a number less than the least root be substituted for x in an = 0. equation, the result will be positive when the equation is of an even degree, and negative when the equation is of an old degree. Again, if h be greater than the greatest root, then every factor will be positive, and consequently the whole product positive. Hence, 2.-1f a number greater than the greatest root be substituted for x in an equation, the result will in all cases be positive. Still again ; suppose h to be at first less than a, but afterward greater than a and less than b. This change in the value of h will change the sign of the factor (h—a), and consequently the sign of the whole product. If in the next place h be made greater than b but less than c, the sign of the product will be changed again, for the same reason as before. And in general, there must be a change of sign every time the variable h passes the value of a real root of the equation, and at no other time. Hence, 3.-If when two numbers are substituted in succession for x, in an equation, the results have contrary signs, there must be at least one real root included between these numbers. It may be observed, also, that if one, three, five, or any odd number of roots be included between the two numbers substituted, the results will show a change of signs. But if an even number of roots be included, there will be no change of signs. 453. If P denote the numerical value of the greatest negative coefficient in an equation, and n the number of terms which precede the first negative coefficient, then "VP+1 will be a superior limit of the positive roots of this equation. Let 20+ 2cm + Axm-' + Bxm-' + Cxm-to...+Tx+U=0. (1) If we omit those positive terms, if any, which occur between 2cm and the first negative term, and then put —P for the coefficient of every other term after um, we shall have " — (P.xm-n+Pam-n-+ +Px+P)=0. (2) Now it is evident that any value which, substituted for x, will give a positive result in eq. (2) will give a positive result also in eq. (1). For, the sum of all the negative terms in (1) can not possibly be greater than the negative part of (2); besides, there may be one or more positive terms in (1) which are omitted in (2). + Dividing every term of (2) by 3r, we obtain Р Р + :0. (3) Make x = "VP+1=r+1, where r is put in the place of ľ for the sake of simplicity. Remembering that P=ym, we have 1+ +...+ (i+1)" (r+1)nti (a +1m-++1) Summing the geometrical series found in the parenthesis, by [360, (B')] the equation becomes zon-1 zon п gon п (=) = 0. n-1 = 0. Now since - +1 r Now since is less than unity, the expression which consti r+1 tutes the first member of (4) is positive. Moreover, the negative te must always be less than unity, whatever be the value of r; hence, no value of r, however great, will render the first member of (4) negative. Thus we have shown that if we substitute for x the quantity "P+1, or any greater value, the result will be positive in equation (3); the result will therefore be positive in eq. (2), and also in eq. (1). Hence, by (452, 2), VP+1 is a superior limit of the positive roots in any equation, which was to be proved. In applying the principle just established, the absolute term must be regarded as the coefficient of xo; and if the equation is incomplete, the deficient terms must be counted, in finding n. It should be observed also, that an equation having no negative term can have no positive roots. For, every positive number substituted for a will render the first member positive. That is, no positive value of x can reduce the first member to zero. EXAMPLES. 1. Find the superior limit of the positive roots of the equation X+5x* +2x414x'—26c+10 = 0. Here n = 3 and P = 26. Hence we have, in whole numbers, "P+1 = 8/26+1 = 4, Ans. 2. Find the superior limit of the positive roots of the equation ** +5.0—25x2—12x+68= 0. Ans. 6. 3. Find the superior limit of the positive roots of the equation x_5.x*—9x+12=0. Ans. 4. 4. Find the superior limit of the positive roots of the equation ++3.2-8 = 0. Ans. 3. 154. To determine the superior limit of the negative roots of an equation, numerically considered, Change the signs of the alternute terms, counting the deficient terms when the equation is incomplete; then apply the preceding rule. For, according to (444), the positive roots in the new equation will be numerically the negative roots in the given equation. EXAMPLES, 1. Find the superior limit of the negative roots of the equation X® ——3x® +5x+7= 0. Ans. 37+1=3, in whole numbers. 2. Find the superior limit of the negative roots of the equation *—15x*—10x+24 = 0. Ans. 5. 3. Find the superior limit of the negative roots of the equation «®--3x+2x* +27x*-4x-1=0. Ans. 4. LIMITING EQUATION. 455. If there be one equation whose roots, taken in the order of their values, are intermediate between the roots of another, the former is said to be the limiting equation of the latter. 156. Any equation being given, its limiting equation may be formed by putting its first derived polynomial equal to zero. If a, b, c,....k, 1 are the roots of the given equation X= 0, and a', b', c',....k' are the roots of the derived polynomial X, = 0, each set being arranged in the order of their values, then we are to show that all these roots, taken together, and arranged in the order of their values, will be as follows : a, a', 1, W, c, c',....k, k', I. In both equations, put x = x'+u, developing the terms, and ar |