then take 2a = 47×2 = 94 for a partial divisor, whence we obtain b = 6, the last figure of the root. We should observe that by simply doubling the 7 in the 87, we may obtain 94, the new trial divisor. From these principles and illustrations, we have the following RULE. I. Point off the given number into periods of two figures each, counting from unit's place toward the left and right. II. Find the greatest square number in the left-hand period, and write its root for the first figure in the root sought; subtract the square number from the left-hand period, and to the remainder bring down the next period for a dividend. III. At the left of the dividend write twice the first figure of the root, for a trial divisor; divide the dividend, exclusive of its righthand figure, by the trial divisor, and write the quotient for a trial figure in the root. IV. Annex the trial figure of the root to the trial divisor for a complete divisor; multiply the complete divisor by the trial figure in the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. V. Take the last complete divisor, doubling its right-hand figure, for a new trial divisor, with which proceed as before, till the work is finished. NOTES. 1. If there is a remainder after all the periods have been brought down, annex periods of ciphers, and continue the operation to as many decimal places as are required. 2. If the denominator of a fraction is not a perfect square, the fraction may be first reduced to a decimal, and its root then taken. 6. What is the square root of 6634.1025? 7. What is the square root of 1812886084? 8. What is the square root of .339889? 9. What is the square root of .00524176? 10. What is the square root of 477 ? 11. What is the square root of 11.09 ? 12. Required the square root of 113001. 1369 13. Required the square root of 200. 14. Required the square root of 215. 15. Required the square root of 54. Ans. 81.45. Ans. 42578. Ans. .583. Ans. .0724. Ans. 21.8403+. Ans. 3.33016+. Ans. 37 Ans. TOOT Ans. 2.3604+. CONTRACTED METHOD. 235. When the required root is a surd, the work may be abridged by the method of contracted decimal multiplication. To insure a correct result, each contracted divisor should contain at least one redundant place—that is, one place more than is necessary to produce the required order of units in the product. This figure should be multiplied mentally, and the tens (increased by 1 when the units are 5 or more) carried to the product of the next figure. To illustrate this principle, let it be required to divide 28337 by 53194, correct to 3 decimal places. In multiplying the first divisor, of which the last figure, 4, is treated as redundant, we say 5 times 4 are 20, and reserve the 2 tens for the next partial product; then, 5 times 9 are 45, and 2 tens added make 47, and we write the unit figure of this result for the first in the contracted product. In multiplying the second divisor, 5319, we have 9×3 53194) 28337 (.5327 26597 5319 1740 1595 532 145 106 53 39 37 = 27; hence there will be 3 tens to carry, because 27 is nearer 30 than 20. The third divisor is 532, one unit being carried to 531 of the preceding divisor, because the rejected figure, 9, is greater than 5. 1. Required the square root of 7.12 correct to six decimal places. We continue the operation as usual until we have obtained the dividend, 1776. At this point we omit the period of ciphers, and consider 533 as the divisor; and in multiplying by 3, the new root figure, we carry the 1 ten from the product of the redundant figure 6, and 1 also from the 8 units in this product, making 1601 for the first contracted product. After this we drop one figure from the right, to form each successive divisor, and thus continue till the work is finished. OPERATION. 12.668333 Ans. 7.120000 4 46 312 276 526 3600 5328 44400 42624 5336 1.776* 1601 It will be observed that there are as many figures in the root thus obtained, as there are in the assumed number. From this illustration, we have the following RULE. I. If necessary, annex periods of ciphers to the given number, and assume as many figures as there are places required in the root; then proceed in the usual manner until all the assumed figures have been brought down. II. Form the next trial divisor as usual, but omit to annex to it the trial figure of the root, reject one figure from the right to form each subsequent divisor, and in multiplying regard the right-hand figure of each contracted divisor as redundant. NOTE.—If a rejected figure is 5 or more, increase the next figure at the left by 1. EXAMPLES. 1. Find the square root of 56 correct to 7 decimal places. Ans. 7.4833147+. 2. Find the square root of 14 correct to 7 decimal places. Ans. 3.7416573+. 3. Find the square root of 18 correct to 4 decimal places. Ans. 4.2426+. 4. Find the square root of 19 correct to 6 decimal places. Ans. 4.358898+. 5. Find the square root of 52.463 correct to 7 decimal places. Ans. 7.24313464-. 6. Find the square root of 7 correct to 8 decimal places. Ans. 2.64575131+. 7. Find the value of 5o correct to 5 decimal places. Ans. 11.18034+. CUBE ROOT OF POLYNOMIALS. 236. We may deduce a rule for extracting the cube root of a polynomial in a manner similar to that pursued in square root, Ly analyzing the combination of terms in the binomial cube. If the binomial, a+b, be cubed, we have a3+3a2b+3ab2+b3. We will now consider how the process may be reversed, and the root extracted from the power. We observe 1st. That the first term of the root may be obtained by taking the cube root of the first term of the power. Thus, Va3 = a. 2d. The second term of the root may be found by dividing the second term of the power by three times the square of the first term of the root. Thus, 3ab3ab. 3d. The last three terms of the power may be factored, and written as follows: (3a2+3ab+b2)b or {3a2+(3a+b)b}b. Thus we see that if to the trial divisor, 3a2, we add a correction, 3ab+b2, or (3a+b)b, the result will be a complete divisor, which multiplied by b, will give the last three terms of the power. Hence, the whole operation of extracting the root, a+b, from the cube, a3+3a2b+3ab3+b3, may be written as follows: Having found a, the first term of the root, we take its cube from the whole expression, and obtain 3a1b+3ab3+b3. Dividing the first term of this remainder by 3a3, we obtain b, the second term of the root. To complete the divisor, we first write the quantity 3a+b; and multiplying this by b, we have 3ab+b, which added to the trial divisor, gives 3a2+3ab+b, the complete divisor. Multiplying this by b, and subtracting the product from the dividend, there is no remainder, and the work is complete. 237. To recapitulate, we may designate the quantities employed in the foregoing operation, as follows: 238. Next, suppose there are three terms in the root, as a+b +c. Assumes a+b; then s+c = a+b+c; and we have (s+c)'s+3s'c+3sc2+c. If we proceed as in the last example, we shall obtain a+b,.or that part of the root represented by s, and subtract its cube from the whole expression. There will then be left 3s3c+3sc2+c3, which may be factored and written (3s2+3sc+c2)c or {3s2+(3s+c)c}c. And we perceive that 3s will be the new trial divisor to obtain c, and that (3s+c)c will be the new correction. The value of 3s2, or 3(a+b)2, may be obtained by multiplication. It will be more convenient, however, to derive it by the addition of three quantities already used in the operation. Thus, |