increase in multiplying the divisor, with the figure thus found annexed to it, by said figure. 1 Place this figure in the quotient, and also at the right of the divisor. Multiply the divisor (thus increased) by this last figure in the quotient, the product will be the subtrahend to place under the resolvend, to be subtracted therefrom. To the remainder bring down another period, and proceed as before, till the work is completed. NOTB. The number of points over the given square, shows how many figures there will be in the root. These examples are proved by multiplying the root by itself, and if any remain take it in, di add it to the product, as in proving division, the result will be equal to the first given number. EXAMPLES. 1. What is the square of 12 ? 12 The number 43 multiplied 19 by itself produces it's square. 144 Ans snovi In this example I find that 1 is the highest number to place in the root ot quotient, as the left hand period is 1, which I place both in the quotient, and under that period, and subtract it from the period, find none re. mains; I then bring down the next period 44, and double the quotient for a divisor, to find a second figure of the root; this divisor 2 I can have twice in 4, (omitting the right hand figure in the resolvend) I set 2 in the quotient or root, and also at the right of the divisor, then I multiply the divisor tbus increased, by this last figure in the root, and set the product under the resolvend, for a subtrahend, and subtract it therefrom; and as there are no more periods to bring down, the work is done ; and as there is no remainder, the root is a perfect one. 4. What is the square of 345 ? root of 119025 ? 119025(345 Ansi 9 64)290 256 3. What is the square 1035-5, Note 1. The Agure found by dividing all but the right hand figure in the resolvend, will sometimes prove 100 great on account of the increase, by multiplying it with the other figures in the divisor ; in which case a less number must be taken, 80 48 the product, may not exceed the resolvend. Note 2. When there is a remainder after extracting the square root from the last resolvend, and a more exact root is required, annex iwo ciphers to the remainder, and proceed as before ; if there be another remainder, annex two ciphers to it, and proceed on to as many places of figures as you please. off 131 senesti NOTE 3. All figures found in the root aften annexing ciphers, will be decimals. In die 5. Required the square root of 1729, i *g'zal til, at luua 1729(41,581+ Ans. 16. 41,581 41,581 381)129 41,581 81 332648 207905 825)4800 41581 4125 166324 Remainder, 20439 X8308)67500 CLX in vels, 66464 men vi Preofg 1799,000000 83161) 103600 83161 20439 Note. Double the square root is the denominator of the remaining fraction, thus, 6. Required the square root of 1296: Abs. 36. 7. What is the square root of 360000 ? Ans. 600 8. What is square foot of 137641 ? Ans. 371. 9. Required the square root of 123692,89. Ans. 351,7. IC. Required the square root of 138387. Ans. 372+. 11. What is the squate toot of 408326 ? Ans. 6397.. The square root is useful in geometry, measuring of superficies, surveying, &c. 12. A surveyor is required to lay out 1000 acres of land, in form of a square, how many rods will be the: kength of each side ? Ans. 400 røde.. Note. The 12 example is solved by reducing the acres to rods, and then extracting the square root of those tods.. 13. There is a ditch 20 feet wide, full of water, on one side of the ditch is a wall is feet hight from the UZ water; it is required to make a ladder that will just reach the top of the wall from the opposite side of the ditch ; how long must it be? Ans. 25 feet. To find the length of the ladder, add the square of the width of the ditch, and the square of the heighth of the wall together, and extract the square root of their sum. * To extract the square root of a vulgar fraction. B Rule. Reduce the fraction to its lowest term, then extract the square roots of the numerator, and also of denominator, and set them down fraction wise, it will be the root required. NOTE. The same rule holds good for extracting the cube or any other root of a vulgar fraction. EXAMPLE rukt: 3. Required the square root of 40% 5)48}(flowest terms. The given number must be prepared for extraction by placing a point over the unit figure, and every third figure each way from it, so as to have periods of three figures each. There will be the same number of figures in the root, as there are periods, either of whole numbers, or of decimals. When there are decimals given, and do not make complete periods of three figures each, annex ciphers sufficient to complete the right hand periud. The first figure sought in the root, is the root of the greatest cube contained in the first or left hand period, and (for discinction) is called a, which is placed in the quotient. The cube of this figure is to be subtracted from the first period. To the remainder bring down the next period for a new resolvend. Then to 300 times the square ofa; add 30 times a, their sum will be a divisor, by which the resolvend is to be divided to find the next figure in the quotient, or root, which next figure is called e. Then to 300 times the square of a xe, add 30 times the square of exa, and also the cube of e; their sum will be a subtrahend, which place under the resolvend, and subtract it therefrom.* To the remainder bring down the next period, which will make another resolvend. All the figures now in the quotient or root (taken together) are called a, with which proceed as before, to find a divisor. With this new divisor find another quotient (or root) figure, which caffe. Proceed with the new a and e as before, to find a subtrahend, which place under the last resolvend, and subtract it theretrom. Τα the remainder bring down the next period, and proceed as before, calling all the figures now in the root a, to find a divisor, and thus proceed till the work is done. 3.1"' 970. EXAMPLES. • Sometimes the figure found by dividing as above will make the subtrahend greater than the resolvend, in which casc a less number must be taken as in the square root. 16083 2001 90119 20 biti |