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514. An Imperfect Power is a number whose root can not be found exactly.

EVOLUTION BY FACTORING.

PROCESS.

515. 1. What is the square root of 1225 ?

ANALYSIS. —Since the 5)1 225

square root of a number 5) 245

is one of its two equal

factors, we may find the 7) 49

square root of 1225 by 7)7 11225=5 X 7=35 separating it into its.

prime factors, and finding the product of the numbers forming one of the two equal sets of factors. The prime factors are 5, 5, 7 and 7, and 5 and 7 form one of the two equal sets. Therefore their product, 35, is the square root of 1225.

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RULE.—Separate the numbers into their prime factors. Arrange these factors into two, three, four, or any number of sets containing the same factors, according as the second, third, fourth, or any root is to be found.

The product of the factors which form a set will be the root.

This method is valuable only when the numbers whose roots are sought are perfect powers.

2. Find the square root of 144. 256. 324. 576.
3. Find the cube root of 64. 512. 4096. 13824.
4. Find the fourth root of 1296. The fifth root of 248832.

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516. 1. How does the number of figures which express the square of units, compare with the number expressing units ?

2. How does the number of figures required to express the cube of units compare with the number expressing units?

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3. Write the numbers, 10, 99, 100, 999, 1000, and under them their second and third powers.

4. How does the number of figures required to express the second powers compare with the number of figures required to express the given numbers?

5. How does the number of figures required to express the third powers compare with the number of figures required to express the given numbers ?

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517. PRINCIPLES.—1. The square of a number is expressed by twice as many figures as the number itself or one less.

2. The cube of a number is expressed by three times as many figures as the number itself or one or two less.

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EXERCISES.

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518. Tell by referring to the principles, how many figures there are in the following:

1. In the square of 21. Of 15. Of 115. Of 4156.
2. In the cube of 19. Of 25. Of 316. Of 6184.
3. In the square of 35. Of 29. Of 584. Of 8196.
4. In the cube of 59. Of 67. Of 999. Of 9999.

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5. How many figures or orders of units are there in a number if the second power of it is expressed by 4 figures? By 7 figures? By 9 figures ?

6. How many figures or orders of units are there in a number if the third power of it is expressed by 6 figures ? By 8 figures ? By 12 figures ? By 11 figurës? - By 21 figures ? By 25 figures ?

519. PRINCIPLES.—1. The orders of units in the square root of a number correspond to the number of periods of two figures each into which the number can be separated, beginning at units.

2. The orders of units in the cube root of a number correspond to the number of periods of three figures each into which the number can be separated, beginning at units.

SQUARE ROOT.

520.

A

1. What is the square root of 576, or what is the side of a square whose area is 576 square units ? 1ST PROCESS.

ANALYSIS.- According to Prin. 576 (20

1, Art. 519, the orders of units 202 400 4

in the square root of any number

may be determined by separating 2 X 20=40) 176 24

the number into periods of two (40+4) X4=176

figures each, beginning at units.

Separating 576 thus, there are 204

found to be two orders of units in the root, or it is composed of tens

and units. Since the square of 20*20* 100

tens is hundreds, 5 hundreds must be the square of at least 2 tens. 2 tens or 20 squared is 400, and 400 subtracted from 576 leaves 176, therefore the root 20 must be in

creased by such an amount as will B

D exhaust the remainder. 20

The square (A) already formed

from the 576 square units is one 20

whose side is 20 units, but inasmuch as the number of units was

not exhausted, such additions must A

be made to the square that they will exhaust the units and keep the figure a square. The necessary

additions are two equal rectangles

B and C, and a small square D. Since the square D is small, the area of the rectangles B and C

20

2

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is nearly 176 units. The area, 176 units, divided by the length of the rectangles, will give the width, which is 4 units. The width of the additions is 4 units, and the entire length, including the small square, is 44 units; therefore the area of all the additions is 4 times !4 units, or 176 square units, which is equal to the entire number jf units to be added. Therefore the side of the square is 24 units, or the square root of the number is 24.

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2D PROCESS.

ANALYSIS.—In the same manner 5.76 (24 as before, the root of this number is

shown to consist of tens and units. t= 22 :4

The tens can not be greater than 2t =40 176

2; therefore we write 2 tens for a 4

partial root. Squaring and sub2t +u ==44 176 tracting, there is a remainder of

176, which is composed of 2 times the tens X units + units?, Art. 506. Since 2 times the tens multiplied by the units is much greater than the units squared, 176 is nearly two times the tens multiplied by the units. Therefore if 176 is divided by twice the tens, or 40, the quotient will be approximately the units of the root. Dividing, the units are found to be 4.

Since the tens are to be multiplied by the units, and the units are to be multiplied by the units or squared, and these results are to be added, to abridge the process the units are added to twice the tens and the sum multiplied by the units. Thus, 40 + 4 is multiplied by 4, making 176. Therefore the square root of 576 is 24.

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When the root consists of more than two orders of units, the process for solving is similar to that already given.

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2D PROCESS.

2. Find the square root of 137641.

1ST PROCESS.

13.76.41 (300

9 00 00 70 300 X 2=600+70=670)4 76 41 1

4 69 00 371 370X2=740+1=741 17 41

7 41

13.76.41(371

9
67)476

469
741)741

741

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In the first process the steps are given with considerable fullness, while in the second process all steps are abbreviated as much as possible.

RULE.Separate the number into periods of two figures each, beginning at units.

Find the greatest square in the left-hand period, and write its root for the first figure of the required root.

Square this root and subtract the result from the left-hand period, and annex to the remainder the next period for a dividend.

Double the root already found for a trial divisor, and by it divide the dividend, disregarding the right-hand figure. The quotient or quotient diminished will be the second figure of the root.

Annex to the trial divisor for a complete divisor, the figure last found, multiply this divisor by the last figure of the root found, subtract the product from the dividend, and to the remainder annex the next period for the next dividend.

Proceed in this manner until all the periods have been used thus. The result will be the square root sought.

1. When the number is not a perfect square, annex periods of ciphers and continue the process.

2. Decimals are pointed off into periods of two figures each, by beginning at tenths and passing to the right.

3. The square root of a common fraction is found by extracting the square root both numerator and denominator separately, or by reducing it to a decimal and then extracting its root.

Extract the square root of the following: 3. 2809.

7. 70756. 4. 3969.

8. 118336. 5. 4356.

9. 674041. 6. 9216.

10. 784996.

11. 938961. 12, 5875776. 13. 12574116. 14. 30858025.

15. Find the value of y/222784; 711390625. 16. Find the value of 1.763876; 1 .30858025.

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