Hence there must be one positive, one negative, and two imaginary roots. Since x2(x2-4)+(x4 − 12) is positive if x = 2, we have a superior limit 2; and similarly obtain the inferior limit 2. Hence we have to find the sign of f(-2), ƒ(− 1), ƒ(0), ƒ(1), and f(2). A simple calculation shows that the required five signs are respectively+−−−+. Therefore one root lies between 2 and 1, another between 1 and 2. EXERCISE 165 Determine the character of the roots of the following equa tions: 1. 2 x22x2+5x-11= 0. 2. x2+3x-3x2-4=0. 3. x-3x3- 12 x − 16 = 0. 4. x3+ax − b = 0, if a and b are positive. 5. 2+ax+b=0, if a and b are positive. Locate the roots of the following equations: 6. 4x3-10 x2 + 2x+3=0. 7. x2+4x2-4x-8=0. 8. x3-6 x2+ 4 x + 10 = 0. 9. x3+4x-5=0. 10. x+2x+2=0. 11. 2-5x-4x+19=0. CHAPTER XXIX SOLUTION OF HIGHER EQUATIONS COMMENSURABLE ROOTS 557. The principal features of the method for finding commensurable roots were discussed in the preceding chapter. There are, however, a few additional propositions, which greatly facilitate the finding of commensurable roots. 558. If r is an integral root of the equation with integral coefficients f(x) = 0, then f(x) must be an integer for any integral x- ጥ value of x, for this quotient is a rational, integral function of x. Since f(1) is easily found, we usually apply this test first for x=1. I.e. f(1) must be an integer for any integral root r. 1 -r Ex. 1. Determine the integral roots of the equation 10x+17-16 x2+2x-20=0. The factors of - 20 are 1, ±2, ±4, ±5, ±10, ± 20. f(1) = 7; that is, 1 is not a root. Rejecting all values of i -r which are not factors of 7, and the corresponding values of r, only 2 remains. As synthetic division shows that 2 is not a root, the equation has no integral roots. Ex. 2. Determine the integral roots of the equation The inferior and superior limits are respectively - 2 and +25. Hence the roots may be 1, +1, 2, ... 20. f(1)=249 +281 + 72 - 180 = 126, hence 1 is not a root. Rejecting all values of 1-r which are not factors of 126, and the corresponding values of r, there remain - 1, 2, 3, 4, 10, and 15. ƒ( − 1) = 2 +49 + 281 − 72 — 180 = 80; hence — 1 is not a root. If r = 2, 3, 4, 19, 15, −1−r=−3, −4, −5, -11, – 16. Rejecting all values of 1-r, which are not factors of 80, and the corresponding values of r, there remain 3, 4, and 15 as possible roots. Synthetic division shows that 3 and 4 are not roots, but 15 is a root. Hence 15 is the only integral root. 559. Newton's method of divisors is very similar to synthetic division; but the numerical work is sometimes simpler, especially when the exponents or the coefficients are very large. Let r be a root of the equation аx2+а1x2-1+An-2 + ··· + An-12+ a1 = 0. ... The right member being an integer, the left member must be an integer. Denoting this integer by 92, dividing by r, and transposing, As before, the left member must be an integer. Denoting it by 93, dividing by r, and transposing, 560. The work can be arranged in a manner very similar to the one used for synthetic division. In fact, the work differs from synthetic division only in two points: (1) We commence at the last term and work toward the first. (2) We divide instead of multiplying. Ex. Determine if 2 is a root of 2x4x3-10 x2 + 15x-14=0. 2-1-10+15 - 14 [2 -2-3+ 4-7 0 −4 - 6+ 8-14 Explanation. Take down - 14, divide it by 2, and add the quotient (-7) to the preceding coefficient. Divide this sum (8) by 2, and add the quotient to the preceding coefficient, etc. As there is no fractional quotient, and as the last sum is zero, 2 is a root. The coefficients of the depressed equation are the numbers in the second line with their signs changed, viz. +2+3−4 +7. 561. If a number is not a root, this fact is discovered as soon as we come to a fractional quotient. This constitutes the advantage of Newton's method over synthetic division. I.e. to determine if 7 is a root of x2+x3- 10 x2+15x-14=0, we have 562. The method for finding commensurable roots may be summarized as follows: 1. Make the coefficients integers, and determine all integers that may be roots. To be roots, integers must satisfy the following conditions: (a) They must be factors of the absolute term. (b) They must lie between the inferior and the superior limits. (c) They must conform with Descartes' Rule. must be an integer, if r is an integral root. Simi f(-1) f(2) -1-2-p etc. 2. Try if the numbers which satisfy (1) are roots of the equa tion. The trial may be made by (a) Synthetic Division. (b) Newton's Method. NOTE. Very small numbers, as +1, or - 1, are more easily tested by direct substitution. E.g. if x = 1, then ƒ(x) = the sum of the coefficients. 3. After all integral roots are removed, the equation has no more commensurable roots, if the first coefficient is unity. 4. If the first coefficient is not unity, transform the equation into one in its simplest form (i.e. first coefficient=1) with integral coefficients, and proceed as before. NOTE. If the first coefficient is very small, it is sometimes convenient to apply this last transformation (4) at the very beginning. Ex. 1. Solve the equation. 2x+5x3-114 x2 - 238 x − 240 = 0. The inferior and superior limits are respectively - 11 and 8. Hence the roots may be ±1, ±2, ±3, ±4, ±5, ±6, −8, — 10. f(1)=2+5 114-238-240 = — 585. Hence 1 is not a root. 2, 3, 4, 3, 6, If r=-10, -8, −6, −3, −4, −3, −2, −1, 1-r= 11, 9, 7, 6, 5, 4, 3, 2, −1, −2, −3, −4, −5. |