know the length of A, but we know that B and C are each 20 rods long, and, of course, both together are 2X20-40 rods long. 40, then, will be sufficiently near to divide by. 176÷40-4 (neglecting remainder.) We suppose 4, then, to be the units' figure, or the side of A. This 4 we add to 40, to get the length of abcd, which is 44. Then, the length, 44X4 the width, 176, the size of the figure a b c d. This agrees with its actual size, and hence we know 4 to be the units' figure of the root. Therefore 24 rods is the length of one side of a court containing 576 square rods. == The above operations show the process here illustrated. The large square, 400, is first subtracted; then the remainder is divided by 40, which is twice 20, the root already found. The quotient found is 4. This is added to 40 making 44, and the sum multiplied by 4 produces 176, which, being subtracted, leaves no remainder. In one of the operations the cyphers are retained at the right of the square number 400, and of its root 20. In the other, these cyphers are dropped, as in common division. When they are dropped, however, a cypher must evidently be understood at the right of the divisor. The intelligent pupil will easily see that when the root has more figures, the process is similar. For the first two periods, it is exactly the same. The third period must then be brought down, the whole root already found doubled for a divisor, (understanding a cypher at the right, as above,) and so on as before. All that is to be observed in this case is, that each divisor must be obtained by doubling the whole root already found. This may be proved by a diagram, as above. Another set of figures like A, B and C must be constructed outside of those. If the pupil does not succeed in the demonstration, it is recommended to the teacher to exhibit it. It is almost needless to remark that we may sometimes have a remainder after the last period is brought down. Thus, in the example above, if the number were 580, instead of 576, 24 would still be the nearest whole number root, and a remainder of 4 would be left. To such a remainder, we may annex periods of cyphers, and continue the root to decimals. Each period, so annexed, must of course contain two cyphers. Likewise, if any dividend is too small to contain the divisor, we must put a cypher in the root, and bring down another period. 2. Extract the square root of 5,499, 025. 3. Find the square root of 2. 2(1.4142+ ROOT. 24/100 281 400 4685/23425 23425 00 2824 11900 411296 28282 60400 56564 3836 In extracting the root of 2, we are obliged to annex periods of cyphers, after * obtaining one figure in the root. Such a root as this, must, of course, be surd; for every dividend ends with a cypher, and the first figure of each subtrahend is the product of some figure by itself, since the last figure of every divisor is the same as the quotient figure by which it is multiplied. But no significant figure, multiplied by itself, produces a product ending with a cypher. Hence, there will always be a remainder, and the root will be, of course, infinite. If, then, there is a remainder, when all the significant figures of any number have been employed in Evolution, the root of that number is a surd. The same is true in case of other roots, as well as of the square. It will be seen, that, as we carry a root to decimals, by annexing periods of cyphers below the units' place, so we should annex periods of significant decimals, if they were in the given number. Hence, decimals are to be pointed off by twos, from the units' place. 4. A farmer wished to lay out a field, in the form of a square, to contain 529 square rods; how long must he have made one side ? A. 23 rods. What is the length of one A. 31 ft. 5. A square floor contains 961 sq. ft. side? 6. Find the sq. root of 784. A. 25. Of 487,204. A. 698. A. 746. Of 441. A. 21. 2,916. A. 54. A. 28. Of 676. A. 26. Of 625. Of 638,401. A. 779. Of 556,516. Of 1,024. A. 32. Of 1,444. A. 38. Of Of 6,241. A. 79. Of 9,801. A. 99. Of 17,956. A. 134. Of 32,761. A. 181. Of 39,601. A. 199. Of 488,601. A. 699. 7. Find the sq. root of 69. A. 8.3066239. Of 97. A. 9.8488578. Of 299. A. 17.2916165. 8996644. Of 282. A. 16.7928556. 351. A. 18.7349940. Of 83. A. 9.1104336. Of 222. A. 14.Of 394. A. 19.8494332. Of Of 979. A. 31.Of 999. A. 31.6069613. Of Of 892. A. 29.. Of 699. A. 26.4386081. 2889757. Of 989. A. 31.4483704. 397. A. 19.9248588. Of 687. A. 26.2106848. 8663690. To find the root of a fraction, take the root both of numerator and denominator, or, if this cannot be done, reduce the fraction to a decimal, and extract its root. The same may be done with mixed numbers. 36 Of 160801? A. 8. What is the sq. root of 25? A. §. 491. Of 237183? A. 187. Of 439339? A. . Of 616225? A. 788. 85 Of ? A. 771. 617796 9. Find the sq. .645497. Of 173. 693 942841 483025 root of 2. A. .8660254. Of 5. A. Of A. .193649167. Of.A. .83205. Of. A. .288617394+ From the above illustrations and examples, we have the rule, 1. HAVING POINTED OFF, SUBTRACT FROM THE HIGHEST PERIOD THE GREATEST SQUARE CONTAINED IN IT, PLACE THE ROOT IN THE QUOTIENT, AND TO THE REMAINDER BRING DOWN THE NEXT PERIOD FOR A DIVIDEND. II. DOUBLE THE ROOT ALREADY FOUND, (UNDERSTANDING A CYPHER AT THE RIGHT,) FOR A DIVISOR, AND DIVIDE THE DIVIDEND BY IT, FOR THE NEXT FIGURE OF THE ROOT. IN III. ANNEX THIS FIGURE TO THE DIVISOR, WHICH, SO CREASED, MULTIPLY BY THE SAME FIGURE FOR A SUBTRAHEnd. IV. SUBTRACT THE SUBTRAHEND FROM THE DIVIDEND, TO THE REMAINDER BRING DOWN THE NEXT PERIOD FOR A NEW DIVIDEND, AND SO PROCEed. The proof is by Involution. NOTE. The roots of many powers may be found by repeated extractions of the square root. Thus, the square root of the square root is the 4th root; the square root of the 4th root the 8th root, and so on. The same may be done by means of the cube root, and by the square and cube roots combined. EXTRACTION OF THE CUBE ROOT. §. CI. By similar reasoning to that in § C. it may be shown, that, when three numbers are multiplied together, the product can never consist of more figures than all the factors together, nor of fewer than the same number less two. Hence, we infer, in like manner, that if any number is pointed off into periods of three figures each, from the units' place, the number of periods will be equal to the number of figures in its cube root. Thus, how many figures in the cube root of 27054036008? Point, thus, 27054036008 There are four periods, and, of course, the root consists of four figures. 1. A cubic solid contains 13,824 cubic feet. How many feet in length is one edge of the solid ? 13,824 is the cube of 24. One side of the solid, then, is 24 feet long. This number, 24, is the same as that used in ex. 1, of the last §. Let it be involved to the cube, the products of the digits composing it, being preserved distinct. In § C. we have already performed the first multiplication, and the several products were 400+80+80+16, represented by distinct diagrams. Multiply by 24 again, and each product may be represented by a solid. 400+80+80+16=576 20+4= 24 Thus, (taking 4, the units' figure,) 16, (=the square, A, § C.)X4= 64 the cube A. 80 (=long figure B, C,) X4-320-the solid B. 80 (figure C, §C,) ×4=320=the solid C. 400 (=square D.) X4 =1,600 the solid D. Then, (taking the 2 tens,=20) 16×20= 320 the solid E. 80x20=1,600-the solid F. 80x20=1,600= the solid G. 80x400-8,000=the cube H. N Now, if the solids A, B, C, and D, should be placed immediately in front of E, F, G, and H, respectively, and the whole brought close together, a cube, I, K, L, M, N, O, would be formed 243-13,824. In this cube, the solid H,-8,000, is the cube of the tens in the root; the solids, B, C, E,=320 each, are products of the tens into the square of the units; the solids, D, F, G,=1,600 each, are products of the units into the square of the tens; and the solid A, 64, is the cube of the units. Let the given number now be pointed off, thus, L The cube of the tens' figure is thousands, and will, the greatest cube to be found in the left hand period. and its root, 20. This cube (the figure H) being 5,824 remains, (=the figures, A, B, C, D, E, Fand G.) E, F and G, are each, on one side, as long as H, (that is, may all be placed side by side, in one solid, thus, b of course, be This is 8,000, taken away, As B, C, D, 20 feet,) they If this The little solid, A, must, for the present, stand by itself. solid, a bedef were divided by its whole upper surface, we should obtain its thickness, a b, which is the units' figure of the root. The upper surfaces of G, F and D, we know, because each surface is equal to 20×20, or the square of the root already found. Neglecting E, C, B, and also the little cube A, then, 20×20×3=1,200, will be sufficiently near for a divisor. 5,824÷1,200 4, (neglecting remainder.) We suppose, then, that 4 is the units' figure, or the thickness, a b, which is equal to the side of A. This 4, we multiply by 202 for each of the solids G, F, D; we then square 4, and mul. tiply it by 20, for each of the solids, E, C, B; and finally cube it, for the solid, A. Then 202 X4X3+42×20×3+43-5,824. This agrees with what we know to be the solidity of A, B, C, D, E, F and G; and hence, we know 4 to be the units' figure of the root. Therefore, 24 is the side of a cube, containing 13,824 solid feet. The above operations show the process here illustrated. The large cube, 8,000, is first subtracted, and the remainder, 5,824, is then divided by 1,200, which is 3 times the square of 20, (the root already found.) We then multiply 3 times 202, or 1,200, by the quotient 4, 3 times 42 by 20, and, finally, cube 4. These three results, added, form the subtrahend. In the second operation above, the cyphers are omitted. It is evident, however, that when a divisor is obtained from the root already found, a cypher must be understood after that root. This, by squaring, will bring two cyphers at the right of the divisor. For the same reason, when a subtrahend is obtained as above directed, the first of the products which compose it, will have two cyphers, the second, one, and the third, none at the right. All these cyphers may be omitted, if they are understood, as in operation 2d, and the numbers arranged for addition, exactly as though the cyphers were expressed. Perhaps it will be better for pupils, at first, to write them out in full, as in operation 1st. When there are more than two periods, the operation is similar. A single period is all that need be brought down at once. Periods of cyphers, (of three places each,) may be annexed at the right, if necessary, and the root car. ried to decimals. In like manner, significant decimals may be poin. ted off towards the right, from the separatrix. It may sometimes happen that the subtrahend found as above, will be larger than the dividend. This occurs, because the divisor is smaller than the whole surface of the solid, abcdef. When this is the case, the quotient, or figure of the root last found, must be diminished, and a new subtrahend found. When no subtrahend can be obtained, smaller than the dividend, a cypher must be placed in the root, and another period brought down. 2. Extract the cube root of 48,228,544. Cyphers 362 3=3888)1572544 Dividend. 362X4X3=15552 omitted. +36X42X3= 1728 =1572544 Subtrahend. +43= 64 00 3. In making an excavation, there were thrown out 616,295,051 solid feet of earth. If it were all formed into a cubic mass, what would be the length of one side? A. 851 ft. 4. A box in the form of a cube, contains 9,261 cubic inches. What is the length of one side? A. 1 ft. 9 in. 5. From a cubical cellar were thrown out 510,082,399 ft. of earth. What was one side of the cellar? A. 799. 6. What is the side of a cubical solid, containing 988,047,936 cubic feet? A. 996 ft. 7. Find the cube root of 941,192,000. A. 980. 958,585,256. A. 986. Of 478,211,768. A. 782. 494,913,671. A. 791. 196,122,941. A. 581. 57,512,456. A. 386. Of Of Of 445,943,744. A. 764. Of Of 204,336,469. A. 589. Of Of 6,751,269. A. 189. Of 39,651,821. A. 341. Of 42,508,549. A. 349. Of 510,082,399. A. 799. Of 469,097,433. A. 777. 8. Find the cube root of 7. A. 1.912933. Of 41. A. 3.448217. Of 49. A. 3.659306, Of 94. A. 4.546836: |