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EXTRACTION OF ANY ROOT.

531. The root corresponding to any perfect power may be obtained by resolving that power into its prime factors, and multiplying together one of each number of equal factors denoted by the exponent of the required root. Thus one of each two equal factors of the power will give the second or square root; one of each three equal factors will give the third or cube root ; one of each four equal factors will give the fourth root; and so on.

532. When the index or exponent of the root to be extracted is a composite number, the root may be obtained by successive extractions of the simpler roots denoted by the several factors of that exponent. Thus the fourth root may be obtained by extracting the square root twice in succession ; the sixth root by extracting the square root and then the cube root; and so on.

EXAMPLES 1. Required the fourth root of 50625 ?

Ans. 15.

BY FACTORS.

BY SUCCESSIVE EXTRACTIONS.

X 5 405

5150625

50625 225 510125

4

42 106 5 2025

84

225/15, Ans.
445 2225
381

1
2225
3 27

25 125
39

125 x 3

0 5 X 3 15, Ans. 2. What is the square root of 998001 ? Ans. 999. 3. What is the cube root of 262144 ?

Ans. 64. 4. What is the fourth root of 43046721 ? 5. What is the fifth root of 14348907 ?

Ans. 27. 6. What is the sixth root of 11390625 ?

Ans. 15. 533.° When the given number is an imperfect power, or otherwise, or the exponent denoting the root is prime, or otherwise, the required root may be found by an elegant process, perfect in principle, called from its inventor,

HORNER'S METHOD.

OPERATION

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Ex. 1. Required the cube root of 92959677. Ans. 453.

The greatest cube 0

92959677 453 contained in the left4 16 64

hand period we find to

be 64, whose root, 4, 4 16 28959

we write as the first 4 32 27125

figure of the required

root. This figure, 4, 4800 1834677

we write under the ci625 1834677

pher of the first column; 120 5425

0 and adding it to the 5 650

cipher obtain 4, which

sum multiplied by the 4 gives 16; and 125

607500 the result, 16, we write under the cipher 5 4059

of the second column, and by addition

obtain 16, which sum, multiplied by the 4, 130 611559 gives 64; and the result, 64, we write in 5

the last column under the left-hand period,

92, of the given number. The 64 sub1350

tracted from the 92 above it gives for a remainder 28. 3 We next add the 4 to the last term, 4, of the first

column, obtaining 8; and the result, 8, multiplied by the 1353

4, gives 32, which we write under the last term, 16, of

the second column, and, adding the same together, obtain 48. We next add the 4 to the last term, 8, of the first column, obtaining 12; and, annexing one cipher to the last term, 12, of the first column, obtaining 120, two ciphers to the last term of the second column, obtaining 4800, and to the remainder in the last column bringing down the next period, 959, obtaining 28959, we complete the work preparatory to the finding of the second root figure.

To determine that root figure, we take the last term, 4800, of the second column, for a trial divisor, and the last term, 28959, of the last column, for a dividend; and, dividing, 6 would appear to be the second figure of the root. This, on trial, however, is found to be too large; we therefore take 5, which answers. This 5 we add to the last term, 120, of the first column, obtaining 125; which sum, 125, multiplied by the 5, gives 625, and that product, added to the last term, 4800, of the second column, gives 5425 ; and this result, 5425, multiplied by the 5, gives 27125, which, written in the last column and subtracted from the figures above it, gives a remainder 1834. Then we add the 5 to the last term, 125, of the first column, obtaining 130 ; and the result, 130, multiplied by the 5, gives 650, which we write under the last terr, 5425, of the second column, and by addition obtain 6075. We next add the 5 to the last term, 130, of the first column, obtaining 135; and, annexing one cipher to the last term, 135, of the first column, obtaining 1350, two ciphers to the last term, 6075, of the second column, obtaining 607500, and to the remainder

in the last column bringing down the next period, 677, obtaining 1834677, the work is completed preparatory to finding the third root figure.

To determine that figure, as before, we divide the last term of the third column by the last term of the second. We thus obtain 3, which, added to the last term of the first column, gives 1353, which sum, multiplied by the 3, gives 4059; and that product, being added to the last term of the second column, gives 611559; and that sum, multiplied by the 3, gives 1834677, which being exactly as large as the last term of the third column, on being written under it and subtracted there is no remainder. The given number is therefore a perfect

power, and the cube root sought is 453. In practice, the work may be performed with less figures, by writing down in the several columns only the results.

RULE. Commence as many columns as there are units in the exponent of the root to be extracted, by writing the given number as the head of the right-hand column, and a cipher as the head of each of the others.

Separate the given number into as many periods as possible of as many figures each as the exponent of the root requires ; and having found the nearest root of the left-hand period, write it as the first figure of the required root.

Write this figure in the first column, and, having added it to what stands above it, multiply the sum by the same figure, and write the product in the second column ; add, in like manner, in the second column, and multiply the sum by the same figure, writing the product in the third column; and so proceed, writing the last product in the last column, and subtracting it from what stands above it.

Then add the same figure to the last term of the first column, multiply the sum by the same figure, and add the product to the last term of the second column; and so on, writing the last product in the last column but one. Repeat the process, stopping each time with one column farther to the left, till the last product shall fall in the second column.

Add the figure found for the rout to the last term of the first column ; annex one cipher to the last number in the first column, two ciphers to the last number in the second column, and so on; and to the last number in the last column bring down the next period for a dividend.

Take the last term of the column next to the last for a trial divisor, and see how often it is contained in the dividend, and write the result as the next figure of the root.

Add this figure to the last term of the first column, multiplying the sum by the same figure, add the product to the second column, and so on; proceed as before, till all the periods have been brought down, or an answer sufficiently exact has been obtained.

NOTE 1. When any dividend will not contain its corresponding trial divisor, write a cipher in the root, bring down to the dividend another period. annex an additional cipher to the last term of the first column, two additional

eiphers to the last term of the second column, and so on; and use the same trial divisor as before, increased however by the additional ciphers.

NOTE 2. — When the given number does not have an exact root, periods of ciphers may be annexed.

NOTE 3. — When the root is required to many places of decimals, the work may be contracted by rejecting one figure at the right from the number in the column next to the last, two from the number in the column next farther to the left, and so on, and otherwise proceeding as directed in the rule, except that the new figure of the root is not added to the first column. As soon as all the figures are rejected in the number of the first column, the remainder of the work may be performed as in contracted division of decimals (Art. 276).

EXAMPLES.

2. Required the fourth root of 1.016397 to eight places of decimals.

Ans. 1.00407427.

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3. What is the cube root of 41673648563 ? Ans. 3467. 4. What is the cube root of 43614208 ?

5. What is the cube root of 1.05 to six places of deci. mals?

Ans. 1.016397. 6. What is the fifth root of 184528125 ?

Ans. 45. 7. Required the fourth root of 100 to six places of deci mals?

Ans. 3.162278. 8. Required the fifth root of the fourth power of 9 to seven places of decimals ?

Ans. 5.7995466.

APPLICATIONS OF POWERS AND ROOTS.

534. A TRIANGLE is a figure having three sides and three angles. When one of the sides of a triangle is perpendicular to another side, the opening between them is called a right angle, and the triangle is called a right-angled triangle.

The lowest side, A B, is called the base of the triangle A B C, the side B C the perpendicular, the longest side, A C, the hypothenuse, and the angle at B is a right angle. Also, the line B C, being perpendicular to the base, is the altitude.

535. The square described upon the hypothenuse of a rightangled triangle is equivalent to the sum of the squares described upon the other two sides.

Perpendic.

Hypothenuse.

A

Base.

Thus, if the hypothenuse AC be 5 feet, the basé A B 4 feet, and the perpendicular BC 3 feet, then 52 : 4? + 32, or 25 16 + 9.

B

» 25

536. To find the hypothenuse, the base and perpendicular being given.

Add together the square of the base and the square of the perpendicular, and extract the square root of the sum.

Thus, if the base be 4 and the perpendicular 3, the hypothenuse will equal v 42 + 32

5. 537. To find the perpendicular, the base and hypothenuse being given.

Subtract the square of the base from the square of the hypothenuse, and extract the square root of the remainder.

Thus, if the base be 4 and the hypothenuse 5, the perpendicular will equal 5— 4 = N9= 3.

538. To find the base, the hypothenuse and perpendicular being given.

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