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218. When the number is a perfect power and the factors are readily found, the root may be obtained as follows:

WRITTEN EXERCISES.

1. Find the square root of 3025, and the cube root of 3375.

PROCESS. 5)3025

3)3375 5)605

3)1125 11)121

3)375 11

5)125

5)25 Hence V 3025 = 5 x 11 = 55, Ans.

5 Hence 3375 = 3 x 5 = 15, Ans.

RULE.

Resolve the number into its prime factors, and take the product of one of every two equal factors for the square root; the product of one of every three equal factors for the cube root, etc.

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SQUARE ROOT. 219. By involution

1 1. 10% = 100. 1002 10000.

92 81. 992 = 9801. 9992 = 998001. The

square of 1, the smallest number of one figure, is 1; the square of 9, the largest number of one figure, is 81 ; hence the square of a number consisting of one figure must contain one figure or two figures.

The square of 10, the smallest number of two figures, is 100; the square of 99, the largest number of two figures, is 9801; hence the square of a number of two figures must contain three figures or four figures—that is, twice two, or twice two less one. The same may

be shown for the square of a number consisting of any number of figures. Hence the

220. Principle.—The square of a number contains twice as many figures as the number itself, or twice as many less one.

WRITTEN EXERCISES. 1. Find the square root of 196.

ILLUSTRATION.

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А

Pointing the number off into periods of two figures each, by ART. 219 we find that there are two figures in its square root; hence the root will consist of tens and units.

The greatest number of tens whose square is contained in 196 is 1 ten, or 10 units. This is shown in the illustration by the square A, whose sides are 10 units and whose area is 10%, or 100 square units. Subtracting 100 from 196, the remainder is 96. This we see consists principally of two rectangles B and C, each of which is 10 units long. Since they nearly complete the square, their area is nearly 96 square units. If we divide 96 by their length, we shall find their width. The length of both rectangles is 2 x 10 20. Dividing 96 by 20, we find the width to be 4 units. Adding the length of the small square D, whose sides are 4 units, the entire length of the remaining surface is 24 units. Multiplying this by the width, 4, the area is 24 x 4 = 96 square units. Subtracting, nothing remains. Hence the square root of 196 is 14.

In practice we omit the ciphers and condense the process, as in the next example. 2. Find the square root of 8464. FULL FORM.

IN PRACTICE. 84'64 90 + 2 = 92 84764 92, Ans. 902 81 00

81 Trial divisor, 2 x 90 180 3 64

182 364 2

364 Complete divisor, 182 3 64 From this example we derive the following

RULE.

Begin at the unit's place and separate the number into periods of two figures each.

Find the largest number whose square is contained in the lefthand period; write it as the first figure of the root, subtract its square from the left-hand period, and annex the next period to the remainder.

Double the root found, for a trial divisor; divide the remainder, omitting the last figure, by this divisor, and annex the quotient to the root, and also to the trial divisor.

Multiply the complete divisor by the second term of the root, and subtract the product from the remainder.

If other periods remain, double the root already found, for the next trial divisor, and proceed as before.

When a cipher occurs in the root, annex 0 to the trial divisor and bring down the next period (see Ex. 3). 3. Find the square root of 254016. PROCESS.

The second trial divisor is 10, and the second 25'40'16 504 25

remainder is 40. Since 10 is not contained in 4, 1004

4016 we place a 0 in the root and annex a 0 to the 4016

trial divisor, and bring down the next period. Find the square root of, o 256. 13. 12321.

22. 315844.
5. 441.
14. 12544.

23. 944784.
6. 625.
15. 41616.

24. 277729.
r. 2025.
16. 104976.

25. 519841.
8. 2704.
17. 262144.

26. 998001.
9. 4489.
18. 368449.

27. 1104601,
10. 7056.
19. 767376.

28. 4016016.
11. 8836.
20. 103041.

29. 5499025.
12. 15129. 21. 180625.

30. 6270016.

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We thus see that the square of a decimal contains twice as many decimal places as the decimal itself; hence to extract the square root of a decimal, we point of the decimal into periods of two figures each, beginning at the decimal point.

31. Find the square root of .6889, and of 33.4084.

PROCESS. 1.68/89/.83, Ans.

33.40'84 5.78, Ans.
64

25
163
489

107

840 489

749 1148 9184

9184

Find the square root of, 32. .0289.

36. 21.16. 33. .4225.

37. 547.56. 34. .6561.

38. .003969. 35. .8464.

39. 1.679616.

40. 204.7761. 41. .00009801. 42. .000001. 43. 1.0201.

SQUARE ROOT OF IMPERFECT SQUARES. 222. When a number is an imperfect square, periods of ciphers may be annexed and the process continued as far as desired. 44. Find the square root of 14 to three places of decimals.

PROCESS.
14.00'00'00 3.741 +, Ans.

9
67 500

469
744 3100

2976 7481 12400

7481

Find the square root of the following numbers to four places of decimals: 45. 2. 47. 5.

49. 7.
46. 3.
48. 6.

50. 8.

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