EXTRACTION OF THE SQUARE ROOT. 220. EXTRACTION of the square root of a number consists in finding a number which, multiplied by itself once, will produce the given number. We have seen (Art. 219) that there are only nine perfect squares which can be expressed by one or two figures, viz. 1, 4, 9, 16, 25, 36, 49, 64, 81, the square roots of which are the first nine numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9. The square root of any number expressed by one or two figures cannot therefore contain more than one figure, that is, units. Should the number be an imperfect square, its root will be found between two whole num. bers, which differ from each other by 1, and whose squares are the next greatest and smallest squares, between which the given number is comprised. Thus, 46 is comprised between 36 and 49, which are the squares of 6 and 7; the square root of 46 is then between 6 and 7. Again, 96 is between 81 and 100, the square roots of which are 9 and 10; the square root of 96 is between 9 and 10. Q. In what does the extraction of the square root consist? How many perfect squares are there which contain only one or two figures? What are their roots? How many figures will the square root of a number comprised of one or two figures contain? Between what numbers will the square root of a number which is not a perfect square be found? Between what numbers will the square root of 96 be found? 45? √37? √24? 221. Every number may be considered as composed of a certain number of tens and a certain number of units. Thus, 16=1 ten+6 units, or 10 + 6, and 98 = = 9 tens +8 units 90+ 8. Now, the square of a number such as 16, will be the product of 10+6 by itself. To see fully of what parts the square will be composed, we will keep the units, tens and hundreds of the square separate, as follows: = Commencing on the right, 6 units of the multiplicand, multiplied by 6 units of the multiplier, give 36 units 62 = square of the units. 1 ten or 10 of the multiplicand, by 6 units of the multiplier, give 6 tens or 6 × 10= product of these tens by the units. Multiplying now OPERATION. 10+6 10+6 10×6+62 102 +10 × 6 102+2(10×6)+62 by the ten of the multiplier, the first product is 6 tens, or 6×10 product of the tens by the units; the next product is 10× 10=102=100=square of the tens. As the same result will be obtained for any other number, we conclude, that the square of a number composed of tens and units contains the square of the tens, plus twice the product of the tens by the units, plus the square of the units. Q. Of what parts may every number be considered as composed? What does, 16 contain? 25? 59? 125? What does the square of a number composed of tens and units contain? are 222. The squares of the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000. 100 is the smallest number composed of three figures, and its square root contains two figures, and is 10; 9999 is the largest number composed of four figures, and its square root is less than 100, being comprised between 99 and 100. Hence, when the given square contains three or four figures, the root will contain two figures; that is, tens and units. Q. What is the square of 10? 20? 40? 60? 90? 100? What is the smallest number composed of three figures? What is the square root of 100? What is the largest number composed of four figures? What is its square root? If a number contain three figures, how many figures will its square root contain? If four figures, how many figures in the root? EXAMPLES. 1. Extract the square root of 4096. Since the number contains four figures, its square root will contain two figures; that is, tens and units. But the square of 1 ten is 100; therefore, the square of the tens of the root cannot be found in the two right-hand figures of the given num OPERATION. 4096 ( 64 36 6x2=124) 49.6 496 Now 4096 is ber. We separate these figures by a dot. contained between the two squares 3600 and 4900. Its root must then be between 60 and 70. The required root therefore contains 6 tens. We place 6 on the right of the given number, from which we separate it by a line. We then subtract its square 36 hundred from 40 hundred, which leaves a remainder 4, to which we bring down the next two figures 96. After subtracting the square of the tens from the given number, the remainder 496 must contain twice the product of the tens by the units, plus the square of the units, (Art. 221). But since tens multiplied by units cannot give a less product than ten, twice the product of the tens by the units cannot be contained in the right-hand figure 6 of the remainder, which is units simply. This product must therefore be found in the two left-hand figures 49. Dividing this number by twice the tens, we get 4 for the units of the required root. Placing 4 in the units' place of the root, and also on the right of the divisor 12, and multiplying 124 by 4, the product 496 is exactly equal to the remainder. The root sought is therefore 64, which may be proved by multiplying 64 by itself, 64× 64-4096. It will be observed that in multiplying 124 by 4, we first multiply 4 by 4, and obtain 16, which is the square of the units of the root, and then 12, which is twice the tens of the root, multiplied by 4, the units, gives the double product of the tens by the units. The product 496 is therefore com posed of the same parts as the remainder after the first operation. 2. Extract the square root of 1444. 8 units × 8 units square of units= Twice tens X units 64 480 544 We separate the two right-hand figures as before. There being again 4 figures in the power, the root contains tens and units. The greatest square in 14 is 9, the root of which is 3. The tens of the root will be 3. Subtracting its square 9 from 14, and bringing down the two other figures 44, we have for a remainder 544. Dividing now the two left-hand figures 54 by 6, which is twice the tens of the root, we find it goes 9 times; but 9 is too large, since 54 contains the tens arising from the square of the units, as well as those produced by the double product of the tens by the units. Trying 8 as the next lower number, and placing 8 on the right of 6, and multiplying 68 by 8, the result 544 is exactly equal to the remainder before found. 39 is therefore the root sought. By examining the operation of multiplying 68 by 8, we see that the square of the units is 64, which contains 6 tens, and these tens are carried into the tens resulting from the multiplication of the double tens by the units. 3. Extract the square root of 729. The given number containing 3 figures, the root will contain tens and units. Separating the two right-hand figures, the greatest square in 7 is 4, whose root is 2. Subtracting the square of 2 from 7, and bringing down 29, the remainder is 329. Dividing 32 by 2x2=4, the quotient is 8; K OPERATION. 729 (27 4 2×2=47) 32.9 329 but the tens produced by multiplying 8 by itself, make the result too large. We find 7 to be the units of the root. Multiplying 47 by 9, the result is 329. 27 is the root sought. 4. Extract the square root of 56169. Since the given number contains more than four figures, the square root will contain more than two. Regarding the numbers as still composed of tens and units, the number 56169 will contain 5616 tens and 9 units. Now, the square of the tens cannot be comprised in the two right-hand figures; we OPERATION, 56169 (237 4 2x2=43) 16.1 23×2=467) 326.9 129 3269 point them off as before, and the square of the tens mi then be found in the remaining figures. But since there are more than two figures left, the tens of the root will contain more than one figure, and we may separate the two righthand figures of 561 as if its root alone had to be extracted; and in like manner we should continue to divide off the given number into periods of two figures from the right, if the part left contained more than one figure. Since 4 is the greatest square contained in 5, the root of which is 2, the first figure in the root will be 2. Subtracting the square of 2 from 5, and bringing down the next period 61, we have 161 for a remainder. Dividing 16 by twice the first figure of the root, that is, by 4, we get 3 as the second figure of the tens of the required root. Placing 3 in the root, and also on the right of the divisor, and multiplying 43 by 3, and subtracting the product as before, we have 32 left. Bringing down the next period 69, and dividing 326 by twice the tens of the root, that is, by 23× 2, we find 7 for the units of the root. Proceeding as before, we find 237 to be the root sought, containing 23 tens and 7 units. It will be observed, that there will be always as many figures in the root as there are periods. From the above examples we may deduce the following |