WRITTEN WORK. annexing the next period to the remainder gives 679, which, for convenience, we call a dividend. This dividend contains the remaining part of the square of the two highest denominations, and must therefore contain the product of two factors (135, m) one of which is the sum of twice the part of the root already found plus the next figure of the root, and the other is the next figure of the root. If we knew the first of these factors, which we may call the TRUE DIVISOR, we could readily find the other by dividing 679 by it. But as we only know the value of the highest denomination, we have to make twice that value, which is the largest part of the true divisor, a TRIAL DIVISOR. Twice 9 18, but as the 9 is tens with reference to the next figure of the root, the 18 must be tens, and in dividing by it we drop the right-hand figure of the dividend. Dividing 67 by 18 gives 3 for a quotient, which must either be equal to or greater than the next figure of the root. * 877969(937 We now complete the true divisor by adding 3 (the 81 last quotient figure) to 18 tens (the trial divisor). The 183) 679 sum is 183, which multiplied by 3 gives 549. This 549 being less than the dividend, shows that 3 is not too large. Hence, 93 represents the two highest deno. 1867) 13069 minations of the root, and, as the root contains three 13069 figures, the 93 must be tens. Subtracting the 549 from 679, and annexing the next period to the remainder, gives 13069 for a new dividend, which contains the remaining part of the square of the tens and units, and therefore must contain the product of two factors, such as described in (135, m). We therefore proceed to form a trial divisor by doubling the 93. Twice 93 is 186, which, as tho 93 is tens, must be tens. Dropping the right-hand figure of the dividend, as before, and dividing the remaining part, 1306, by 186, gives 7 for a quotient, which is either equal to or larger than the units figure of the root. We now complete the true divisor by adding 7 (the last figure found) to 186 tens (the trial divisor). The result is 1867, which multiplied by 7 gives 13069. This being equal to the dividend, shows that 7 is the true units figure of the root. Hence, the required root is 937. (a.) These principles are of universal application, and furnish the following rules – 1st. Divide the given number into periods of two figures each, beginning with the units. The left-hand period may contain one or two figures. * Since the trial divisor is less than the true divisor, this figure can never be too small. 2d. Find the greatest square in the left-hand period, and place its root as the first figure of the required root. 3d. Subtract the square thus found from the left-hand period, and to the remainder bring down the next period, calling the result a dividend. 4th. Double the part of the root already found for a trial divisor. 5th. See how many times this trial divisor is contained in all of the dividend, excepting the right-hand figure, and write the quotient as the next figure of the root, and also place it at the right of the trial divisor, to form a true divisor. 6th. Multiply this true divisor by the root figure last found, and subtract the product from the dividend. 7th. Bring down the next period to right of the remainder, to form the next dividend. 8th. Double the part of the root already found, for a trial divisor, and proceed as with the last trial divisor. (b.) Find the square root of each of the following numbers 2. 529 8. 105625 9. 47524 10. 401956 11. 863041 12. 31684 13. 654481 14. 10323369 137. Square Root of Fractions. (a.) To square a fraction, we multiply it by itself, i. e. we square its numerator for a new numerator and its denominator for a new denominator. (b.) Hence, to extract the square root of a fraction, we must extract the square root of its numerator for a new numerator, and the square root of its denominator for a new denominator. ILLUSTRATIONS. -VÝ g V.25 = ,5 V.0081 = .09 (c.) If both terms are not perfect squares, only an approximate value of the square root can be obtained. (d.) If the denominator is not a perfect square, both terms should be multiplied by such a number as will make it so. This number may be either the denominator or the product of all the prime numbers which are found as factors in the denominator, 1, 3, 5, or any other odd number of times. (e.) In order that the denominator of a DECIMAL FRACTION may be a perfect square, its numerator must contain an even number of decimal places. ILLUSTRATION. — The denominators of .04, 3.27, and .1951 are perfect squares, but the denominators of .4, .327, and .00625 are not. (f.) Hence, to extract the square root of a decimal fraction, annex a zero, if necessary to make its number of decimal places Then extract the root as in whole numbers, observing that there will be one decimal place in the root for every two places in the given fraction. even. (g.) The root may be found to any number of decimal places by annexing two zeroes for every additional figure desired in the root. ILLUSTRATION. – V.9 V.90 V.9000 V.900000, etc. (h.) What is the square root to 5 decimal places of 10. .5625 ? 11. .0025 ? 12. .1429? 13. 88? 16. 86.53 ? 138. Relation of Cube to Root. 1=1 10% = 1,000. 43 = 64 73 = 343 125 88 = 512 63 = 216 93 - 729 10003 1,000,000,000. 10000o = 1,000,000,000,000. (a.) The above table shows, first, that there are but nine entire numbers below 1000 which are perfect cubes. (b.) Second, that the entire part of the cube root of any number below 1000 will be less than 10, and will therefore contain but one figure; of any number between 1000 nd 1000000 will lie between 10 and 100, and will therefore contain two figures, or, in other words (c.) The entire part of the cube root of a number containing 1, 2, or 3 figures will contain one figure; the entire part of the cube root of a number containing 4, 5, or 6 figures will contain two figures, and so on with the larger numbers. (d.) Hence, if we should begin at the right of any number and separate it into periods of three figures each, the number of periods will be the same as the number of figures in the entire part of the cube root. The cube of the highest denomination of the root will be found in the left-hand period. The cube of the two highest denominations will be found in the two left-hand periods, etc. (e.) We have next to consider how the cube of a number consisting of two parts is formed from those parts. To do this, let a represent any number whatever, and b any other number. Then a + b will represent the sum, and (a + b)* the cube of the sum of any two numbers whatever. (f.) But (a + b)' = a + 2ab + v'. Multiplying this by a gives a' + 2 a' b + a b', and multiplying it by b gives a' b + 2 a b + bo. Adding these results together gives a + 3 a 6+ 3 a b' + @o, as in the following written work. a' + 2 a b + b = (a + b)' atb a' + 2 a' b + a ba = a x a’ + 2ab + b a' b + 2 a b + bo: b x a + 2 ab + 6 a' + 3 a' b + 3 a 6° +6= (a + b) x (a + 2ab + b)=(a + b)* (g.) But the last three terms 3 ao b + 3 a 6° + 5* = 3 a' times b + 3 a b times b + Bo times b (3 a' + 3 a b + 6) x b. But the last two terms within the parenthesis 3 ab + b=3 a times b + b times b (3 a + b) x b. Hence (3 ao + 3 a b + 6) x 6 = [3 a' + (3 a + b) 6] x b and (a + b)' = a + 3 a 6 + 3 a +b=a' + [3 a' + (3 a + b) b] x b. (h.) Subtracting a' from this leaves [3 a + (3 a + b) 6] x 6, which shows that (i.) If from the cube of the sum of two numbers we subtract the cube of the first number, the remaiņder is the product of two factors, one of which, b, is the second number, and the other, 3 ao + (3 a + b) x b, is the sum obtained by adding to 3 times the square of the first number, the product of 3 times the first number plus the second multiplied by the second. 139. Method of extracting the Cube Root. 1. What is the cube root of 279726264 ? Solution.— This number contains three periods of three figures each, which shows that there are three figures in its root. Moreover, the greatest cube in 279 is the cube of the highest denomination of the root, and the greatest cube in 279726 is the cube of the two highest denominations of the root. Hence, we may find the first two figures of the required root by find. ing the cube root of 279726 as though it were units. The greatest cube below 279 is 216, the root of which, 6, is the first figure of the required root. Subtracting 216 from 279, and annexing the next period to the remainder, gives 63726, which for convenience we call a dividend. This dividend contains the remaining part of the cube of the two highest denominations of the root, and must therefore (see 138, i) contain the pro |