1. Required the square root of 7.12 correct to six decimal

places.

We continue the operation as usual until we have obtained the dividend, 1776. At this point we omit the period of ciphers, and consider 533 as the divisor; and in multiplying by 3, the new root figure, we carry the 1 ten from the product of the redundant figure 6, and 1 also from the 8 units in this product, making 1601 for the first contracted product. After this we drop one figure from the right, to form each successive divisor, and thus continue till the work is finished.

It will be observed that the number of places in the root is equal to the number of places assumed in the power.

From this illustration, we have the following

RULE.-I. If necessary, annex periods of ciphers to the given number, and assume as many figures as there are places required in the root; then proceed in the usual manner until all the assumed figures have been brought down.

II. Form the next trial divisor as usual, but omit to annex to it the trial figure of the root, reject one figure from the right to form each subsequent divisor, and in multiplying regard the right-hand figure of each contracted divisor as redundant.

NOTE.-If a rejected figure is 5 or more, increase the next figure at the left by 1.

EXAMPLES.

1. Find the square root of 56 correct to 7 decimal places.

Ans. 7.4833147+. 2. Find the square root of 14 correct to 7 decimal places.

Ans. 3.7416573+.

3. Find the square root of 18 correct to 4 decimal places. Ans. 4.2426+.

4. Find the square root of 19 correct to 6 decimal places.

Ans. 4.358898+.

5. Find the square root of 52.463 correct to 7 decimal places.

Ans. 7.2431346+.

6. Find the square root of 7 correct to 8 decimal places. Ans. 2.64575131+.

7. Find the value of 5% correct to 5 decimal places.

Ans. 11.18034-.

CUBE ROOT OF POLYNOMIALS.

236. We may deduce a rule for extracting the cube root of a polynomial in a manner similar to that pursued in square root, by analyzing the combination of terms in the binomial cube. If the binomial, a + b, be cubed, we have

a3 + 3a2b+3al2 + b3.

We will now consider how the process may be reversed, and the root extracted from the power. We observe

1st. That the first term of the root may be obtained by taking the cube root of the first term of the power.

Thus,

2d. The second term of the root may be found by dividing the second term of the power by three times the square of the first term of the root. Thus,

3a2b3a2 = b.

3d. The last three terms of the power may be factored, and written as follows:

(3a2 + 3ab + b2)b or {3a2 + (3a + b) b} b.

Thus we see that if to the trial divisor, 3a2, we add a correction, 3ab+b2, or (3a+b) b, the result will be a complete divisor, which multiplied by b, will give the last three terms of the power.

Hence, the whole operation of extracting the root, a+b, from the cube, a3 + 3a2b + 3ab2 + b3, may be written as follows:

Having found a, the first term of the root, we take its cube from the whole expression, and obtain 3ab+3ab2+b3. Dividing the first term of this remainder by 3a2, we obtain b, the second term of the root. To complete the divisor, we first write the quantity 3a + b; and multiplying this by b, we have 3ab + b2, which added to the trial divisor, gives 3a2 + 3ab + b2, the complete divisor. Multiplying this by b, and subtracting the product from the dividend, there is no remainder, and the work is complete.

237. To recapitulate, we may designate the quantities em ployed in the foregoing operation, as follows:

238. Next, suppose there are three terms in the root, as

a + b + c.

Assume s

a + b; then s+ c = a+b+c; and we have

(s + c)3 = 83 + 38°c + 3sc2 + c3.

If we proceed as in the last example, we shall obtain a + b, or that part of the root represented by s, and subtract its cube from the whole expression. There will then be left 38°c + 3sc2 + c3, which may be factored and written

(3s2 + 3sc + c2)c_or {3s2 + (3s + c)

And we perceive that 3s2 will be the new trial divisor to obtain c, and that (3s + c) c will be the new correction.

The value of 3s2, or 3 (a + b)2, may be obtained by multiplication. It will be more convenient, however, to derive it by the addition of three quantities already used in the operation. Thus,

Last complete divisor,

3a2 + 3ab + b2

Last correction,

3ab + b2

(b).

b2

Square of last term of the root,

382 3 (a + b)2 = 3a2 + 6ab + 3b2.

Let it now be required to find the cube root of the polynomial x6+3x5 3x11x3 + 6x2 + 12x 8.

Having arranged the polynomial according to the exponents of 2, we proceed as in the former example, and obtain a2, the first term of the root, 3x3-3x2-11x3+6x2+12x−8 the first remainder, 324 the trial divisor, and a the second term of the root. To complete the trial divisor according to formula (a), we write three times the first term of the root plus the second, or 3x2+x, for the first factor of the correction. Whence we have (3x2+x)x, or 3x3+x2, for the correction; 34+3x+x2 for the complete divisor; (324+3x3 + x2)x, or 3x5 + 3x + x3, for the product; and - 6x4 12x+6x2 + 12x 8 for the new dividend.

To form the new trial divisor according to formula (b), we have (324+3x+x2) + (3x3+x2) + x2 = 3x1+6x+322; whence, by division, we obtain 2 for the third term of the root. To complete the new trial divisor, we have for the first factor of the correction, 3 (x2+x) − 2 = 3x2 +3x-2. This may be obtained in the operation from the former factor 3+, by simply multiplying its second term by 3, and annexing the 2. We now find the correction, complete divisor, and product as before, and the work is finished. It is evident that three or more terms of the root will sustain the same relation to the next succeeding term, that the first sustains to the second, or the first and second to the third.

239. From the foregoing analysis we derive the following

RULE.-I. Arrange the polynomial according to the powers of some letter, and write the cube root of the first term for the first term of the root; subtract the cube of the root thus found from the polynomial, and arrange the remainder for a dividend.

II. At the left of the dividend write three times the square of the root already found, for a trial divisor; divide the first term of the dividend by this divisor, and write the quotient for the next term of the root.

III. To three times the first term of the root annex the last term, and write the result at the left, and one line below, the trial divisor; multiply this result by the last term of the root, for a correction of the trial divisor; add the correction, and the result will be the complete divisor.

IV. Multiply the complete divisor by the last term of the root, subtract the product from the dividend, and arrange the remainder for a new dividend.

V. Add together the last complete divisor, the last correction, and the square of the last term of the root, for a new trial divisor; and by division obtain another term of the root.

VI. Take the first factor of the last correction with its last term multiplied by 3, and annex to it the last term of the root, for the first factor of the new correction; with which proceed as before, till the work is finished.

1. What is the cube root of 27a3 + 108a2 + 144a + 64?

4. What is the cube root of ao + 9a3b + 24a1b2 + 9a3b3—24a2l1 +9ab5 — b6 ?

5. What is the cube root of a9 234a1 + 257a3 174a2 + 60a

Ans. a2 + 3ab — b2.

6a8+27a-74a6 + 159a5

8? Ans. a3 2a2 + 5a - 2.

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