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6. Find a divisor as before, by doubling the figures already in the root; and from these find the next figure of the root, as in the last article; and so on through all the periods to the last.

NOTE 1. When the root is to be extracted to a great number of places, the work may be much abbreviated thus: having proceeded in the extraction by the common method till you have found one more than half the required number of figures in the root, the rest may be found by divide ing the last remainder by its corresponding divisor, annexing a cypher to every dividual, as in division of decimals; or rather, without annexing cyphers, by omitting continually the right hand figure of the divisor, after the manner of contraction in division of decimals.

NOTE 2. By means of the square root we readily find the fourth root, or the eighth root, or the sixteenth root, &c. that is, the root of any power, whose index is some power of the number 2; namely, by extracting so often the square root, as is denoted by that power of 2; that is, twice for the fourth root, thrice for the eighth root, and

so on.

TO EXTRACT THE SQUARE ROOT OF A VULGAR FRACTION.

RULE.

First prepare all vulgar fractions by reducing them to their least terms, both for this and all other roots.

Then

1. Take the root of the numerator and that of the denominator for the respective terms of the root required. And this is the best way, if the denominator be a complete power. But if not, then

2. Multiply the numerator and denominator together; take the root of the product: this root, being made the numerator to the denominator of the given fraction, or the denominator

denominator to the numerator of it, will form the frac tional root required.

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And this rule, will serve, whether the root be finite or infinite.

Or 3. Reduce the vulgar fraction to a decimal, and extract its root.

EXAMPLES.

1. Required the square root of 5499025.

5499025(2345 the root,

4

43 149
31129

4642090
4 1856

4685|23425
123425

2. Required the square root of 1842,

184-2000(13.57 the root,

I

23 84
3169

265|1520
* 51325

2707/19500
18949

551 remainder.

3. Required

3. Required the square root of 2 to 12 places.

2(141421356237 + root.

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1. Having divided the given number into periods of 3 figures, find the nearest less cube to the first period by the table of powers or trial; set its root in the quotient, and subtract the said cube from the first period; to the remainder bring down the second period, and call this the resolvend.

2. To three times the square of the root, just found, add three times the root itself, setting this one place more to the right than the former, and call this sum the divisor. Then divide the resolvend, wanting the last figure, by the divisor, for the next figure of the root, which annex to the former; calling this last figure e, and the part of the root before found call a.

3. Add

* The reason of pointing the given number, as directed in the rule, is obvious from Cor. 2, to the Lemma made use of in demonstrating the square root; and the rest of the operation will be best understood from the following analytical process :

Suppose N, the given number, to consist of two periods, and let the figures in the root be denoted by a and b.

Then

159

3. Add together these three products, namely, thrice the square of a multiplied by e, thrice a multiplied by the square of e, and the cube of e, setting each of them one place more to the right hand than the former, and call the sum the subtrahend; which must not exceed the resolvend; and if it do, then make the last figure e less, and repeat the operation for finding the subtrahend.

14. From the resolvend take the subtrahend, and to the remainder join the next period of the given number for a new resolvend; to which form a new divisor from the whole root now found; and thence another figure of the root, as before, &c.

EXAMPLES.

Then a+ba3+3a2b‡zab2+b3N given number, and to find the cube root of N is the same as to find the cube root of ■3+3a2b+3ab2+b3; the method of doing which is as follows: á3 +3a2b+3ab2+b3 (a+b root.

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3a2b+3ab2+b3 resolvend,

3a2

+3a

3a +3a divisor.

3a2b

+3ab2

+63

3a2b+3ab2+b3 subtrahend.

*

And in the same manner may the root of a quantity, consisting

of any number of periods whatever, be found.

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