After obtaining the third trial divisor, we commence rejecting one figure from the trial divisor, two from the number at the foot of the preceding column, three from the third column, &c., and proceed in a similar way with each subsequent trial divisor, until the figures from the preceding columns are entirely cancelled. But in every instance, allowance must be made for the product of the figures rejected, as in simple contracted division.

49. Extract the square root of 287; of 5.

50. Extract the cube root of 11; of 25; of 693.

51. Extract the 4th root of 13; of 1.8; of 27.

52. Extract the 5th root of 797.9341.

53. Extract the 5th root of 1.0843.

54. Extract the cube root of 997641.285.

55. Extract the cube root of 1.64193.

56. Extract the square root of 24976819542.42.

57. Extract the 7th root of 27.91.

The SQUARE ROOT of any number may be expressed in the form of a continued fraction, after part of the root is found,—by making each numerator equal to the remainder, and each denominator equal to twice the root found. Thus in extracting the square root of 17, the first root figure is 4, and the remainder 1. Then the true root is 4 + the continued fraction

nearly, giving the first approximate root 32. Second,

nearly, giving a second approximate root 317. Third,

5

6205
276

23'

31

nearly, giving a third approximate root 323. This approximation is of use in affording convenient fractional expressions for those roots which are of most frequent occurrence. Thus, the diagonal of a square is to its side as √2 is to 1. By the rule just given, we obtain successively for approximate values of √2,

The last of these values, 128 or 98, is a very convenient one.

The following is a general rule for the approximation of ANY ROOT desired.

RULE.

Call the first two figures of the root found in the usual way, the

ASCERTAINED ROOT.

Involve the ascertained root to the given power, and multiply by the index of the root for a dividend.

Subtract the power of the ascertained root from the corresponding

periods of the given number, for a divisor. Divide, and reserve the quotient.

To 6 times the reserved quotient, add the index of the root, plus 1, for a second dividend.

To 6 times the reserved quotient, add 4 times the index of the root, subtract 2 from the sum, and multiply by the reserved quotient for a second divisor. Divide, add 1 to the quotient, and multiply by the ascertained root for the true root nearly. If greater accuracy is desired, repeat the process with the root thus found.

By this rule, the number of figures in surd roots, may generally be tripled at each operation.

The following is the application of the rule, in extracting the 5th root of 659901.

2689120122077 = 22.02806, reserved quotient.

reserved quotient 22.02806

6

132.16836

index + 1. 6.

second dividend 138.16836

6 x by reserved quotient =132.16836

4 X 5-2: = 18.

correct to the fourth decimal place.

This contraction is of use in extracting the higher roots. Any root below the 10th may be obtained in the usual way, nearly as readily, and with much greater accuracy.

58. Find convenient fractional approximations to √3.

59. What are the approximate fractional values of √5? 60. Extract the 13th root of 1.08.

61. Extract the 17th root of 1.004.

62. Extract the 100th root of 1.07.

63. Extract the 45th root of 1.2.

64. Reduce 13 to a continued fraction.

65. What are the approximate values of the continued fraction which is equivalent to 27?

CHAPTER XXIV.

ANALYSIS.

ALL the operations of Arithmetic have for their object, the discovery of one or more unknown quantities; and the great difficulty in complicated questions, is to perceive the application of the simple rules which will lead to this discovery.

The examination of any question, in order to determine the relation of the different quantities to each other, is called ANALYSIS. To keep the unknown terms more constantly in view, letters are frequently employed to represent them, and the work expressed in the statement of the question, is performed on these letters, as if their value was known, and we were proving the truth of the answer.

EXAMPLE FOR THE BOARD.

There is a fish whose head weighs 9 pounds; his tail weighs as much as his head and half his body; and his body weighs as much as his head and tail both. What is the weight of the fish?

Now, if 18 added to the half of x gives x, 18 must itself be equal to

2

, and twice 18, or 36, to x. Having found the value of x, we easily obtain the remaining values.

The pupil may analyse the following examples, either with or without the aid of letters.

1. A father's age is 7 times that of his son, and the sum of their ages is 40. What is the age of each?

2. In a certain school there are 45 scholars, and there are twice as many boys as girls. Required the number of each ?

3. A man performed a journey of 135 miles, going twice as far the second day as on the first, and three times as far the third day as on the second. How far did he travel each day?

4. A., B., and C., entered into partnership, contributing in the whole, $4833. B. paid twice as much as A., and C. paid twice as much as A. and B. How much did each contribute?

5. In a certain school of 70 scholars, three times as many