TO EXTRACT THE SQUARE ROOT. * Divide the given number into periods of two figures each, by setting a point over the place of units, another over the place of hundreds, and so on, over every second figure, both to the left hand in integers, and to the right in decimals. Find the greatest square in the first period on the left-hand, and set its root on the right-hand of the given number, after the manner of a quotient figure in Division. * The reason for separating the figures of the dividend into periods or portions of two places each, is, that the square of any single figure never consists of more than two places ; the square of a number of two figures, of not more than four places, and so on. So that there will be as many figures in the root as the given aumber contains periods so divided or parted off. And the reason of the several steps in the operation appears from the algebraic form of the square number of terms, whether two or three or more. Thus, (a + b)? = a2 + 2ab + b2 = u? + (2a + b) b, the square of two terms ; where it appears that a is the first term of the root, and b the second term; also a the first divisor, and the new divisor is 2a + b, or double the first term increased by the second. And hence the manner of extraction is thus: Ist divisor a ) a2 + Zub + b2 ( a + b the root. a? of any 2d divisor 2a + b | 2ab + 62 b2ab + 67 Again, for a root of three parts, a, b, c, thus : (a + b + c) = a + 2ab + b2 + 2uc + 2bc + c = u+ (2a + b) b + (2a + 2b + c) c, the square of three terms, where a is the first term of the root, b the second, and c the third term ; also a the first divisor, 2a + b the second, and 2a + 2b + c the third, each consisting of the double of the root increased by the next term of the same. And the mode of extraction is thus : Ist divisor a) a2 + 2ab +62 + 2ac + 2bc + c? (a + b + c the root. a? Subtract the square thus found from the said period, and to the remainder annex the two figures of the next following period, for a dividend. Double the root above mentioned for a divisor ; and find how often it is contained in the said dividend, exclusive of its right-hand figure; and set that quotient figure both in the quotient and divisor. Multiply the whole augmented divisor by this last quotient figure, and subtract the product from the said dividend, bringing down to it the next period of the given number, for a new dividend. Repeat the same process over again, viz. find another new divisor, by doubling all the figures now found in the root; from which, and the last dividend, find the next figure of the root as before ; and so on through all the periods, to the last. Note, The best way of doubling the root, to form the new divisors, is by adding the last figure always to the last divisor, as appears in the following examples.-Also, after the figures belonging to the given number are all exhausted, the opera tion may be continued into decimals at pleasure, by adding any number of periods of ciphers, two in each period. NOTE, When the root is to be extracted to many places of figures, the work may be considerably shortened, thus: Having proceeded in the extraction after the common method, till there be found half the required number of figures in the root, or one figure more; then, for the rest, divide the last remainder by its corresponding divisor, after the manner of the third contraction in Division of Decimals; thus, 2. To find the root of 2 to nine places of figures. 2 ( 1.41421356 the root. 28294) 3836 ( 1356 1008 2 RULES FOR THE SQUARE ROOTS OF VULGAR FRACTIONS AND MIXED NUMBERS. FIRST prepare all vulgar fractions, by reducing them to their least terms, both for this and all other roots. Then 1. Take the root of the numerator and of the denominator for the respective terms of the root required. And this is the best way if the denominator be a complete power : but if it be not, then 2. Multiply the numerator and denominator together; take the root of the product: this root being made the numeG2 ratcr rator to the denominator of the given fraction, or made the denominator to the numerator of it, will form the fractional root required. Vab b Vab And this rule will serve, whether the root be finite or infinite, 3. Or reduce the vulgar fraction to a decimal, and extract its root, 4. Mixed numbers may be either reduced to improper fractions, and extracted by the first or second rule, or the vulgar fraction may be reduced to a decimal, then joined to the integer, and the root of the whole extracted. EXAMPLES Ans. Ans. 2. 1. What is the root of ? 2. What is the root of 27,? 3. What is the root of it! Ans. 0.866025. 4. What is the root of ? Ans. 0.645497, 5. What is the root of 17? Ans. 4:168333. By means of the square root also may readily be found the 4th root, or the 8th root, or the 16th root, &c, that is, the root of any power whose index is soine power of the number 2; namely, by extracting so often the square root as is denoted by that power of 2; that is, two extractions for the 4th root, three for the 8th root, and so on. So, to find the 4th root of the number 21035.8, extract the square root two times as follows: TO EXTRACT THE CUBE ROOT. I. By the Common Rule*. 1. HAVING divided the given number into periods of three figures each, (by setting a point over the place of units, and also over every third figure, from thence, to the left hand in whole numbers, and to the right in decimals), find the nearest less cube to the first period, set its root in the quotient, and subtract the said cube from the first period; to the remainder bring down the second period, and call this the resolvend. 2. To three times the square of the root, just found, add three times the root itself, setting this one place more to the right than the former, and call this sum the divisor. Then divide the resolvend, wanting the last figure, by the divisor, for the next figure of the root, which annex to the former ; calling this last figure 6, and the part of the root before found let be called . 3. Add all together these three products, namely, thrice a square multiplied by ė; thrice a multiplied by e square, and e cube, setting each of them one place more to the right than the former, and call the sum the subtrahend; which must not exceed the resolvend; but if it does, then make the last figure e less, and repeat the operation for finding the subtrahend, till it be less than the resolvend. 4. From the resolvend take the subtrahend, and to the remainder join the next period of the given number for a new resolvend; to which form a new divisor from the whole root now found; and from thence another figure of the root, as directed in Article 2, and so on. * The reason for poir the given number into periods of three figures each, is because the cube of one'figure never amounts to more than three places. And, for a similar reason, a given number is pointed into periods of four figures for the 4th root, of five figures for the 5th root, and so on. And the reason for the other parts of the rule depends on the algebraic formation of a cube : for, if the root consist of the two parts a + b, then its cube is as follows: (a + b)3 = Q3 + 3a2b+ 3ab2 + b3; where a is the root of the first part a3 ; the resolvend is 30b + 3ab? b3, which is also the same as the three parts of the subtrahend; also the divisor is 30% + 3a, by which dividing ihe first two terms of the resolvend 3u2b + ao”, gives b for the second part of the root; and so on. EXAMPLE |