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10. What is the square root of .000729 ? Ans. 027. 11. What is the square root of 17.3056? Ans. 4.16. 12. What is the square root of 52 16?

Ans. 74 13. What is the square root of 9515?

Ans. 9 14. What is the square root of 36336T?

Ans. 1915. 15. How much is v1.96 ?

Ans. 1.4. 16. How much is 65611?

Ans. 81. 17. How much is 93 ?

Ans. 27. 18. How much is gi ?

Ans. 64. 19. How much is one of the two equal factors of 9645192360241 ?

Ans. 3105671.

526.

When the square root is to be extracted to many places of figures, the work may be contracted thus :

Having found in the usual way one more than half of the root figures required, the rest may be found by dividing the last remainder, with a single figure annexed instead of two, by the last divisor, and proceediny as in contracted division of decimals. (Art. 276.)

EXAMPLES.

1. What is the square root of 785 to five places of decimals ?

Ans. 28.01785.

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The nature and extent of the contraction will be seen by comparing the contracted method with the common method.

2. Extract the square root of 63 to four places of decimals.

Ans. 2.5298+. 3. Required the square root of 2 to five places of decimals.

Ans. 1.41421+. 4. Required the square root of 3.15 to eight places of decimals.

Ans. 1.77482393+. 5. Required the square root of 373 to seven places of decimals.

Ans. 19.3132079+. 6. Extract the square root of 8.93 to eight places of decimals.

Ans. 2.98831055+.

EXTRACTION OF THE CUBE ROOT.

527. The extraction of the cube root of a number is the process of finding one of its three equal factors; or, of finding a factor which, being multiplied into itself twice, will produce the given number.

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528. The common method of extracting the cube root depends upon the following principles:

1. The cube of any number has, at most, only three times as many figures as its root, and, at least, only two less than three times as many. For the cube of a number of a single figure consists of, at most, three figures, and, at least, two less than that number, as 13 1, and 93 729; the cube of a number of two figures consists of, at most, six figures, and, at least, two figures less than that number, as 103 : 1000, and 993 970299;. and so on. Therefore, when a cube number consists of one, two, or three figures, its root will consist of one figure; when of four, five, or six figures, its root will consist of two figures, and so on; and if a number be separated into as many periods as possible of three figures, each commencing at the right, to these periods respectively will correspond the units, tens, hundreds, &c. of the cube root of that number.

2. The cube of a number consisting of TENS and UNITS is equal to the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. Thus, if the tens of a number be denoted by a, and the units by b, the cube of the number a +6

=

36

will be denoted by (a + b = a + 3 a b + 3 ab? + 69. Then, by this formula, if a = 3, and b equal 6, we have 3 tens + 6 units = 30+ 6 = 36, and 363 = (30 + 6) = 303 + 3 (30% X 6) + 3 (30 X 6') + 6 = 46656. Or, analytically, a+b = 30 + 6

36 = 30 + 6

36 (a+b) Xa=302 + 30 x 6

1080 (a+b) xb= 30 X 6+ 62

216 (a+)2 = 30% + 2 X 30 X 6+62

1296 a+b

= 30 + 6 (a + b)2 Xa= 30+2 X 302 X 6+ 30 X 62

38880 (a+b)xb=

302 X 6+2 X 30 X 6+6= 7776 (a+b)3 = 30% + 3 (302 X 6) +3 (30 x 6?)+63= 46656

It is evident, as evolution is the reverse of involution, that from this process of obtaining a cube may be deduced a method of extracting the cube root. Since the cube of a + b is a3 + 3 a b + 3 a b + b, the cube root of a + 3 ab+ 3 ab? + 63 must be a + b. Now.a, the first term of the cube root, is the cube root of a', the first term of the cube; and if a' be subtracted, there will remain 3 a2 b + 3 a b + b, from which b, the second term of the root, is to be obtained. But 3 a’b + 3 a b? + 63 is the same as (3 a2 + 3 ab + b) X b; therefore the remainder equals (3 a' + 3 ab + b) X b. But as 3 ab, three times the tens into the units, plus 6, the

square of the units, is generally much less than 3 a’, three times the square of the tens, we consider that 3 aX b is about equal to the whole remainder, and taking 3 a? (which we know) as the trial divisor, we obtain b, the units. But as the true divisor is 3 a' + 3 ab + b we add three times the tens by the units plus the square of the units, and multiply the sum by the units, which gives a product equal the whole remainder, or 3 a' b + 3 a b + 6.

Since every number of more than one figure may be considered as composed of tens and units, we may have tens and units of units, tens and units of tens, tens and units of hundreds, &c. Hence, the principle just explained applies equally whether the root contains two or more than two figures.

529.

To extract the cube or third root of numbers.

Ex. 1.

What is the cube root of 46656 ?

Ans. 36.

OPERATION.

=

over

PROOF.

Beginning at

46650 36 the right, we sep33

27

arate the given Trial div., 3 X 302 2700

number into pe

19656 3 x 30 x 6 540

riods, by placing 0? 36

a point over the

units figure and True divisor, 3276 x 6 19656

the third figure to the left. Since the number

of periods is two, 36 X 36 X 36 46656

the root will con

sist of two figures, tens and units. Then 46656 = the cube of tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. The cube of tens is thousands, and must therefore be found in the thousands of the number. The greatest number of tens whose cube does not exceed 46 thousands is 3, which we write as the tens figure of the root. We then subtract the 27 thousands, the cube of the 3 tens, from the 46 thousands, and there remain 19 thousands; and, annexing the next period, we have as the entire remainder, 19656, equal three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units, or the product of three times the square of the tens, plus three times the tens into the units, plus the square of the units, multiplied by the units. By dividing this remainder by three times the square of the tens of the root, we obtain the units, or a number somewhat too large. Although it may be too large, it cannot be too small, since the remainder 19656 contains not only three times the square of the tens into the units, but three times the tens into the square of the units, plus the cube of the units. We therefore make three times the square of the tens of the root, 27 hundreds, a trial divisor, with which we divide the 196 hundreds of the remainder, disregarding the 56 units, since they cannot form any part of the product of the square of the tens by the units. The quotient figure obtained, 7, must be the units figure of the root, or a number somewhat larger.

But on undertaking to complete the divisor on the supposition that 7 is the true units figure of the root, we find a divisor too large for the remainder. We therefore take 6, a number one less, and to determine whether it expresses the real number of units in the root, we add to the 27 hundreds of the trial divisor three times the 3 tens of the root into the 6 units, plus the square of the 6 units; and multiplying the true divisor, 3276, thus formed, by the units, and subtracting the product, 19656, from the remainder, there is nothing left. Ilence, 46656 is a perfect cube, and 36 its cube root.

2. What is the cube root of 12326391 ?

Ans. 231.

FIRST OPERATION.

1 2 3 2 6 3 9 i 2 3 1 23 =

8 Trial divisor, 3 X 202 -= 1200

4 3 2 6 3 X 20 X 3 180

32 9 True divisor,

1389 X 3 41 67 Trial divisor, 3 x 230o = 158700

15 9 3 91
3 x 230 X 1 690
12

1 True divisor,

159391 X 1= 15 9 3 91 Since there are three periods the root will contain three figures, the first two of which may be considered as tens and units of tens. Aj the cube of tens cannot give less than thousands, we must find that cube in the two left-hand periods; and as we have tens and units of tens, their cube will equal the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. This we apply to the first two periods exactly the same as when the root consists of but two figures, and thus take from the given number the cube of the 23 tens, which leaves a remainder of 159391. We now consider the given number, 159391, as the cube of a number consisting of 23 tens and a certain number of units, which cube will of course equal the cube of the tens, plus three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units. But the cube of the tens, or (23)”, has already been taken from the given number, leaving a remainder, 159391, which must equal three times the square of the tens into the units, plus three times the tens into the square of the units, plus the cube of the units; or equal the product of three times the square of 23, plus three times 23 into the units, plus the square of the units, multiplied by the units. From this we readily obtain the units, just as when we had but two figures in the required root.

1 2 3 2 6 3 9 i 231

8 Trial divisor, 1200

SECOND OPERATION.

4 3 2 6
63 x 3 189
True divisor, 1389 X 3 =

41 67
32
9

15 9 3 91 Trial divisor,

= 1587 0 0 691 x 1

691 True divisor, 15 9 391 X 1 = 15 9 3 91

23 =

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