243

ARITHMETIC.

II . To the remainder annex the’second period for a dividend,stnd on the leit of it write double the root alrcjtdy found, after themanner of a divisor.

HI. Consider what figure must be annexed to the divisor, sothat, if the result be multiplied by it, the product may be equalto, or the next less than, the dividend, and it w ill be the next figureof the root.

IV. Subtract the product from the dividend, and to the re-mainder bring down the next period, for a new dividend.

V. Find a divisor as before, by doubting the figures alreadyin the root; and from these find the next figures of the root, as inthe last article ; and so on through all the periods to the last.

When the root is to contain many places of figures, havingfound about half of them in the common way, the rest may befound, by dividing the last remainder by its corresponding divisor,bringing down one figure or cypher as in common division, or itmav be done after the manner of contractions in division ofdecimals.

Examplh. Extract the square root of 328 to 12 or 13places.

32S.00(18.11077027627 + root required.

1

28

228

8

224

361

400

1

361

3621

1

3900

3621

332207

3622147

7

279000ft

2335449

23455100

25355029

362215,4 | 100071(027627-)%27628227310O£83

(82.) To Extract thf. Cube Root.

Rule. I. Divide the number into periods of three figures, andfrom the first period, take the nearest cube, which is not greaterthan it; putting its root for the first figure of the required root,and to the remainder bring down another period fora dividend.

II. Multiply the square of the figure found by 3, for a divisor,anti trv it in the dividend, neglecting the units and tens, for ano-ther figure of the root.

III. To three times the first root, annex this trial figure, amimultinlv the result by it, placing the product below’ the divisortwo figures farther to the right, and add these two lines togetherfor the complete divisor ; then multiply by the trial figure, sub-tract the product, and bring down the next period.

IV. Place the square of the figure last found below the completedivisor, and add it, and the two lines above it, for a new divisor,ami try as before, for the next figure of the root.

V. To the number before-mentioned, with the annexed figure,add double that annexed figure, to this sum, annex the last trialfigure, multiply the result by it, and add it to complete the divisoras before ; then repeat the w ork till the whole is finished.

When the root is to be extracted to a great number of places,having found about half of them by the rule, the rest mav befound by proceeding with the last remainder, and last completedivisor, after the manner -of contractions, in division of decimals.

Example.

Required the cube root of 21035.8.

21035.800(27.6049105397

1?

8

07 x 7 =

469

--

14

1609

1303.8

11683

49

•

--

1352800

2187

816 x 6 =

4S90

12

22359600

1341576

36

*'11224000000

22852800

82804x4 =

331216

8

2285611216

9142444864

16

. - 2081555136000

22S594244S

828129X 9= 7453161

-— 2057415281649

228601697961 --

24139854351

127968

13667

2237

ISO

20

The above is an improvement of the rule given in the appendixof Mr. Ingram’s edition of Dr. Ilutton’s Arithmetic, and in itspresent form preferable to any yet published, as will appear bycomparing this example, of which the intire work is given.

(83.) To Extract any Root whatever.

Let G be the given power or number, n the index of thepower, A the assumed power, r its root, R the required rootof G.

Then as the sum of n 1 times A and n — 1 times G, is tothe sum of 1 times G and n — 1 times A, so is the assumedroot r, to the required root of R.

Or, as half the said sum of n -f 1 times A and n — 1 times G,is to the difference between the given and assumed powers, sois the assumed root r, to the dilierence between the true andassumed roots: which difference, added or subtracted, gives thetrue root nearly. 6

That is, n + 1. A + 4— 1. G: n+ 1. G + 77Ti. A :: r ; R

Or, 7!■ -j- 1. A -f- n — 1. J G : A 1 2 * * * 5 G :: r : R to r.

And the operation may be repeated as often as we please, byusing always the last found root for the assumed root, and its nthpower for the assumed power A.

A otc. 1 he assumed power A is to be found by trials nearlyequal to the given power.

Ex. To extract the 5th root of 21035.8.

Here it appears that the vth root is between 7-3 and 7 4.Taking 7'3, its 5th power is 20730'71593. Hence then we findG = 21035’S ; r = 7-3 ; n = 5 ; 4 j/’+T = 3; 1 n — { = 2,

A = 20730-716

G —A = 305-084

A = 20730-716 G = 21035-8

3 2

3 A = 62192-148 420/76

2 G = 42071-6

As 104263-7: 305-084 : : 7-3 : -02136057-39152522135588

104263-7 )’I’227-1132(-0213605 the ditff.

14184 7-3 =)- add

3758 7-321360 = II the root true to the last

630 figure.

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