--

and annex it to 8, the first term in the trial divisor; multiply the result by 3x and subtract the product from the remainder, leaving a second remainder, 16 x 2. 20 x3 10 x1. 4x+x6. Double the root already found, obtaining 8-6x, the first part of the second trial divisor; divide 16 x by 8, obtaining + 2x for the third term in the root; annex this term to the trial divisor 8 6x, multiply the sum 8 - 6x+2x2 by 2 x2 and subtract the product from the second remainder; this operation gives a third remainder, - 8x3+ 6 x1 — 4x3 + x®. Double the root already found, obtaining 8-6x+4x2 for the first part of the third trial divisor; then divide -8 x3 by 8, obtaining 3 as the fourth term of the root. Now from the above process there is no remainder on the completion of the last step. The operation is completed in this case.

This problem could have been solved with the same ease by arranging the expression with respect to the descending powers of x. II. Find the square root of

9 x2-6xb+30 xc +6 xd +b2.

10bc2bd+ 25 c2 + 10 cd + d2.

Arrange the terms with respect to x thus:

9x2-6xb+30xc+6xd+b2-10bc-2bd+25c2+10cd+d2 3x-b+5c+d

It will be noticed that each trial divisor is equal to the preceding with the last term doubled.

287. The fourth root of an expression can be found by extracting the square root of the square root. Similarly, the eighth root may be found by extracting the square root three successive times; and the sixteenth root by four successive extractions of the square root, and so on (see (282).

For example, find the fourth root of

81 x1 432 x3 + 864 x2 768x + 256.

By proceeding as in the foregoing example the square root of the proposed expression will be found to be 9 x2 24x16; and the square root of this is 3 x 4, which is the fourth root of the given

expression.

17.

y2

x2

18.

x2

2 a2 + a2b2 + 2 + 2 b2 +

y2 (2-x)-2 y (1-x2) + x2.

x2 (2—y2) — 2 + 2y2 + F.

19. a2 + 2 ab + 2 ac + b2 + 2bc + c2.

20. 1+ 2x + 3 x2 + 2 x3 + x1.

21. 1+ 4x + 10 x2 + 20 x3 + 25 x1 + 24x3 + 16 xo.

23.

+2 + e22 + 2 €2 + 2 + e-22.

24. 12 gr + 3 q2x2 + 4 q3μ3 + 3 q*p* + 2 q3μ3 + qoμ·o ̧

25. 1+ 2+ 3 e2a + 4 esa + 5 etа + 4 e5 +3 ebа + 2e7а + c&d.

34.

12x 4(n-1)*n* + 4 (n - 1)2x+2(n-2)*n*

-4(n-1) (n − 2)2 + (n − 2).

38.

p2z® — 2p3z2 + p1z° + 2 y ( 1 − 2) z 1 — 2 y( p − y)z3

41. Prove by extracting the root that:

√1 + x = 1 + {x− } x2+1'6 x3-1 § 5x++ x} qx3¬z } } qx®+ ··

How can one of these roots be derived from the other?

42. Calculate to seven decimals the values of 10 by means of one of the series in 41. Here 1/10

=

1 101

and put x =

in 41. 2.

43. Calculate in a similar manner to five decimal places

Here v 2=31 ́1—1.

44. Calculate to six decimal places 11.

Here 11 13011-100.

45. Calculate likewise 3, 15, and 16 to four decimal places.

46. Calculate to five decimal places 1/7 and 1/13. Find the fourth root of:

47. x3 + 4x2 + 6 x1 + 4 x2 + 1.

-1

72x3n-1 144x3n-2 96x3-354x2 + 216 x2n-1 +216x2-2 108.x"

216"-1+81.

THE SQUARE ROOT OF ARITHMETICAL NUMBERS

288. The rule for finding the square root of an algebraic expression makes it possible to derive a rule for finding the square root of numbers in Arithmetic.

The square root of 100 is 10, of 10000 is 100, of 1000000 is 1000, and so on; therefore the square root of a number less than 100 consists of one figure, of a number between 100 and 10000 consists of two figures, of a number between 10000 and 1000000 consists of three figures, and so on. If, therefore, a dot be placed over the figure in units' place of a number equal to or greater than 1, and over every alternate figure, the number of dots will be equal to the number of figures in the root of the number. Thus, the square root of 4096 consists of two figures, the square root of 611524 of three figures, and so on.

289. Find the square root of 4489.

Point the number according to the rule; hence the root will consist of two figures.

4489 3600

60+ 7

120+7 889

889

Let ab denote the root; then a may be taken as the value of the figure in tens' place, and the figure in units' place. Then a is the greatest multiple of 10 whose square is less than 4400; this is found to be 60. Subtract a2 or 3600 from the given number and the remainder is 889. is, by 120, and the quotient is 7, which is the value of b. Hence (2 a+b)b, which is (120+7)7, or 889, is the number to be subtracted. Therefore, since there will be no remainder, the conclusion is that the required root is 67. The ciphers may be omitted for the sake of brevity and the following rule be derived from the process:

Divide the remainder by 2 a, that

4489 36 127 889 889

67

Point off the number into periods of two figures each, beginning with units' place.

Find the greatest number whose square is contained in the first period; this will be the first figure of the root; subtract its square from the first period and to the remainder bring down the next period. Drop the right hand figure of the remainder and divide the number so found by twice that part of the root already found. Annex this quotient to the part of the root already found and also to the trial divisor. Then multiply the divisor as it now stands by the figure of the root last found and subtract the product from the last remainder.

If there are more periods to be brought down the operation must be repeated.

290. Extract the square roots of 481636 and 11566801.

NOTE. The student should note the occurrence of the cipher in the root.

291. If the square root of a number has decimal places, the number itself will have twice as many.

Thus, if 23 is the square root of some number, this number will be (.23)2 = .0529; and if .113 is the square root of some number, the number will be (.113)2 = .012769.

Therefore, there is an even number of places in a decimal which is a perfect square, and the number of decimal places in the root will be half as many as in the given number itself. Hence this rule for extracting the square root of a decimal may be deduced:

Place a dot over the figure in units' place, and over every alternate figure, continuing to the left and to the right of it; now proceed as in the extraction of the square root of whole numbers, and mark off as many decimal places in the result as there are periods in the decimal part of the given number.