REMARK.-The operations in the example of the table, are all performed according to the directions of the rule; but more decimal places have been used in the dividends and divisors, in the latter part of the work, than were necessary. Had we admitted rejected all other but three places of decimals in the dividends, and places to the right, as fast as they occurred, we should still have had the root equally true to at least four places of decimals. But since the figures of the root are decimals, it follows that if the number of decimal places in the dividend does not exceed three, the decimal places in the corresponding divisor should not exceed two; and for every succeeding figure found in the root, one place may be struck off from the right of the divisor. After finding a certain number of figures of the root, it will occur that the numbers to be added to the divisors will fall among the rejected figures, after which the remaining figures of the root will be found by simple division. It should be observed, however, that when places are rejected from the divisor, that whatever would have been carried had the complete multiplication been performed, is still to be carried to increase the last figure retained; and whenever the left-hand figure of those rejected, either in the dividend or divisor, exceeds 4, the last figure retained is to be increased by 1. The following is the last example, wrought on the principle of admitting but three places of decimals into the dividends. The rejected figures, both in the dividends and divisors, are placed a little to the right. 2. Find one root of the equation x3 + x2 = 500. This equation is the same as x3 + x2+0x = 500; hence b=1 c = 0, and N = 500. The first figure of the root found by trial, is 7. 0 56 56 49 161 13.56 500 7.61727975, &c., = x. 392 108 104.736 3.264 174.56 188.48 188.7181 .0001 188.9563 189.1232 189.290 .189.295 1.887181 1.376819 1.323862 .052957 . 037859 . 013251 .001847 .001704 .000143 .000010 .000009 We may always commence REMARK. This example is wrought rejecting the places from the dividends, Ans. x 14.954, &c. 4. Fu e root of the equation x3 + 2x2 + 3x = 5. Find one root o equation x3 + 2x2 13089030. Ans. x 235. 23x = 70. Ans. x5.1345, &c. only one root has been REMARK. In the preceding solut obtained, yet the others may be found with al facility, by finding by trial the first figure in each and then preceding by the rule already given. There is, however, a shorter met for determining the remaining roots.. Subtract the root found, taken with a contrary sign, from the coefficient of the second term of the given equation, and denote the remainder by a. Divide the absolute term by the root found, and denote the quotient by b; then will the roots of the equation x2 + ax + b = 0 be the two remaining roots of the given equation. Method of resolving Higher Equations. 345. The general method of resolving cubic equations, has been explained in Art. 342. We shall now add from Young's Algebra, the method of resolving equations of a higher degree. It has not been thought best to give the general investigation, but merely · to add, for the solution of an equation of any degree, the following general RULE. I. Transpose the absolute term to the second member of the equation. Then, beginning with the co-efficient of the first term, arrange the coefficients of the first member in a row, placing the absolute term to the right. II. Having found the first figure of the root, multiply it by the first co-efficient and add the product to the second co-efficient; then multiply the sum by the same figure of the root and add the product to the third co-efficient; and so on to the last co-efficient: the last sum will be the first divisor, which multiply by the figure of the root and subtract the product from the absolute term: the result will be the second dividend. III. Perform the same operations on the first co-efficient and the set of sums found, as was performed with the co-efficients, and the last sum will be the TRIAL DIVISOR for the second figure of the root. Then perform the same operations on the first co-efficient and the second set of sums, only stop in the column of the last co-efficient but one. Repeat the same operation on the first co-efficient and the last set of sums, but stop in the next left-hand column, and so on until you stop in the column of the second co-efficient. IV. Then find from the trial divisor the second figure of the root, taking care that it be not too large. Take this second figure, and perform with it on the first co-efficient and the last set of sums the same operations as were performed on the co-efficients with the first figure of the root and the sum; in the last column will be the second divisor, which multiply by the second figure of the root and subtract the product from the second dividend. V. The next trial divisor, the next figure of the root, and the true divisor, are found by the principles already explained, and the places of figures in the root may be carried as far as necessary. REMARK. The work, in the example, has been contracted by omitting or cutting off decimal places, as in the operations for the cube root, and in equations higher than the third degree, the con tractions may be begun after the first decimal place of the root is found. By using one period of four decimal places, the root has been found to eight places of figures. Another period of four places, that is, by beginning the contractions later, we should have found four additional places, or x = 9.88600270094. THE END. L |