Square Root of Numbers. 200. The preceding rule is applicable to the extraction of the square root of numbers; for every number may be regarded as an algebraic polynomial, or as composed of a certain number of units, tens, hundreds, etc. Thus If, then, 841 is the square of a number composed of tens and units, it must contain the square of the tens, plus twice the product of the tens by the units, plus the square of the units. But these three terms are blended together in 841, and hence arises the peculiar difficulty in determining its root. The following principles will, however, enable us to separate these terms, and thus detect the root. 201. 1st. For every two figures of the square there will be one figure in the root, and also one for any odd figure. Thus The smallest number consisting of two figures is 10, and its square is the smallest number of three figures. The smallest number of three figures is 100, and its square is the smallest number of five figures, and so on. Therefore the square root of every number composed of one or two figures will contain one figure; the square root of every number composed of three or four figures will contain two figures; of a number from five to six figures will contain three figures, and so on. Hence, if we divide the number into periods of two figures, commencing at the units' place, the number of periods will indicate the number of figures in the square root. 202. 2d. The first figure of the root will be the square root of the greatest square number contained in the first period on the left. For the square of tens can give no significant figure in the first right hand period, the square of hundreds can give no figure in the first two periods on the right, and the square of the highest figure in the root can give no figure except in the first period on the left. Let it be required to extract the square root of 5329. 53,29 (70+3, the root. 49 00 140+3) 429 429 This number contains two periods, indicating that there will be two places in the root. Let a+b denote the root, where a is the value of the figure in the tens' place, and b of that in the units' place. Then a must be the greatest multiple of 10, which has its square less than 5300; this is found to be 70. Subtract a2, that is the square of 70, from the given number, and the remainder is 429, which must be equal to (2a+b)b. Divide this remainder by 2a, that is hy 140, and the quotient is 3, which is the value of b. Completing the divisor, we have 2a+b=143; whence (2a+b)b, that is 143 × 3, or 429, is the quantity to be subtracted; and as there is now no remainder, we conclude that 70+3, or 73, is the required square root. For the sake of brevity, the ciphers may be omitted, provided we retain the proper local values of the figures. If the root consists of three places of figures, let a represent the hundreds, and b the tens; then, having obtained a and b as before, let the hundreds and tens together be considered as a new value of a, and find a new value of b for the units. Required the square root of 568516. Having found 75, the square root of the greatest square number contained in 56,85,16 (754 49 725 1504) 6016 the first two periods, we bring down the 145) 785 6016 203. Hence, for the extraction of the square root of num bers, we derive the following RULE. 1st. Separate the given number into periods of two figures each, beginning from the units' place. 2d. Find the greatest number whose square is contained in the left-hand period; this is the first figure of the required root. Subtract its square from the first period, and to the remainder bring down the second period for a dividend. 3d. Double the root already found for a divisor, and find how many times it is contained in the dividend, exclusive of its righthand figure; annex the result both to the root and the divisor. 4th. Multiply the divisor thus increased by the last figure of the root, subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 5th. Double the whole root now found for a new divisor, and proceed as before, continuing the operation until all the periods are brought down. In applying the preceding rule, it may happen that the product of the complete divisor by the last figure of the root is greater than the dividend. This indicates that the last figure. of the root was taken too large, and this happens because the divisor is at first incomplete—that is, is too small. In such a case, we must diminish the last figure of the root by unity until we obtain a product which is not greater than the dividend. EXAMPLES. 1. What is the square root of 294849? Ans. 543. 204. Square Root of Fractions. We have seen that the root of a fraction is equal to the root of its numerator di vided by the root of its denominator. Hence the square root 23 is or 2.3. So, also, 18.6624 may be written 10' 432 10000' its square root is or 4.32. That is, the square root of a decimal fraction, or of a whole number followed by a decimal fraction, may be found in the same manner as that of a whole number, if we divide it into periods commencing with the deci mal point. In the extraction of the square root of an integer, if there is still a remainder after we have obtained the units' figure of the root, it indicates that the proposed number has not an exact square root. We We may, if we please, proceed with the approximation to any desired extent by supposing a decimal point at the end of the proposed number, and annexing any even number of ciphers, and continuing the operation. We thus obtain a decimal part to be added to the integral part already found. So, also, if a decimal number has no exact square root, we may annex ciphers and proceed with the approximation to any desired extent. 4. What is the square root of 9.878449 ? 5. What is the square root of 58.614336? 6. What is the square root of .558009 ? 7. What is the square root of .03478225? Find the square roots of the following numbers to five decimal places. 205. We already know that the cube of a+b is a3+3a2b +3ab2+b3. If, then, the cube were given, and we were required to find its root, it might be done by the following method. When the terms are arranged according to the powers of one letter, a, we at once know, from the first term, a3, that a must be one term of the root. If, then, we subtract its cube from the proposed polynomial, we obtain the remainder 3a2b+3ab2+b3, which must furnish the second term of the root. Now this remainder may be put under the form (3a2+3ab+b2)b; whence it appears that we shall find the second term of the root if we divide the remainder by 3a2+3ab+b2. But, as this second term is supposed to be unknown, the divisor can not be completed. Nevertheless, we know the first term, 3a2, that is thrice the square of the first term already found, and by means of this we can find the other part, b; viz., by dividing the first term of the remainder by 3a2. We then complete the divisor by adding to it 3ab+b2. If this complete divisor be multiplied by b, it will give the last three terms of the power. Let it be required to find the cube root of 8a3+36a2b+54al2 +2763. 8a3+36a2b+54ab2+27b3 (2a+3b 8a3 12a2+18ab+9b2 ) 36a2b+54ab2+2713 36a2b+54ab2+27b3 Having found the first term of the root, 2a, and subtracted its cube, we divide the first term of the remainder, 36a2b, by three times the square of 2a, that is 12a2, and we obtain 36 for the second term of the root. We then complete the divisor by adding to it three times the product of the two terms of |