9 x1 y2 + 12 x2 y2 + 4 yồ ( 3 x2 y + 2 y3

The given quantity being arranged according to the different powers of x, as in Division, the root of the first term is found to be 3 x2 y, which is put in the quotient's place.

The second power of 3 x2 y is next subtracted from the first term of the given quantity, 9 x4 y2, when there is no remainder; which proves that the true root of this term has been found.

To ascertain the next term of the root, we must divide the second term of the given power by twice the root already found; therefore, the remaining terms of the power, 12 x2 y1 + 4 yo, are brought down for a new dividend.

That portion of the root which has been found, 3 x2 y, is then doubled, and put in the divisor's place. x2

The first term of the new dividend, 12 x2 y1, is di vided by 6 x2 y, and the quotient is found to be 2 y3, which is the second term of the root.

To ascertain whether 3 x2 y2 y3 be the true root required, we should involve it, and subtract its second power from the given quantity, when there must be no remainder. But the second power of the first term, 3 x2 y, has already been subtracted. Now, in the formula, (a + b)2 = a2 + 2 a b + b2, twice the first term of the root, multiplied by the last term,

is equal to the second term of the power; also the last term of the root, multiplied by itself, gives the last term of the power. That is, if we add the last term of the root to twice the first term, and multiply their sum by the last term, the product will be equal to the second and third terms of the power. Therefore, 2 y3 is added to 6 x2 y, to complete the divisor.

The whole divisor, 6 x2 y2 y3, is then multiplied by 2 y3; and the product is subtracted from the new dividend. There being no remainder, we are certain that 3 x2 y2 y3 is the true root required.

From this investigation may be derived the follow ing RULE for extracting the square root of a com.. pound quantity.

1. Arrange all the terms of the given quantity according to the powers of one of the letters, so that the highest power shall stand first, the next highest next, and the rest in order, as in Division.

2. Find the root of the first term, and put it in the place of a quotient.

3. Subtract the square of this root from the first term of the given quantity, and bring down the remaining terms for a dividend.

4. Double the root already found, and put it in the divisor's place.

5. See how often this divisor is contained in the first term of the new dividend, and annex the quotient both to the root already found and to the divisor.

6. Multiply the divisor, thus increased, by the term of the root last found, and subtract the product from the dividend,

7. Double the whole root for a new divisor, and divide as before.

8. Proceed in this way, until the entire root of the given quantity is extracted.

2. What is the square root of 9 x1 — 12 x3 † 16 xa -8x+4?

9x412 x316x28x+4(3x2-2x+2.

In this example, the second term of the root, 2 x, is, because it is the quotient of 12 a3 divided by +62. The second divisor, 6 x24 x, is obtained, as before, by doubling the whole of the root already found.

3. What is the square root of a2 + 2 a b + b2 — 2ac2bc + c2?

a2+2ab+b2—2 ac—2bc+c2 (a + b—s.

4 What is the square root of 4 x2 + 12 x y + 9 y3? 5. Extract the square root of

4 α x3 + x2.

4x3+6x2 4 x + 1?

7. Required the square root of

x6 + 12 x5 + 5x4. - 2 x3 + 7 x2

2 x + 1.

4 α x + x2?

2 x3 + x2.

10. What is the square root of

x2 + 4x22 y + 4 y2 — 4 x2 — 8y + 4? 11. Extract the square root of

Ga1 b2 - 30 a2 b + 12 a1 ba + 25 — 20 a2 3 + a4 4 a4 b6.

12. What is the square root of

4 + 4 a2 b + a1 b2 — 12 y2 — 6 a2 b y2 + 9 y1?

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13. Required the square root of

4 a2 4 a b + b2 + 12 a c 6 b c + 9 c2.

14. What is the square root of

x+4x5+ 10 xa + 20 x3 + 25 x2 + 24 x +- 16? 15. Find the second root of

4 at 4 at 12 a3 + a2 — 6 a + 9.

16. Extract the square root of

a2+2ab+b2

c2-2cd+d2

Extract the square root of the numerator and of the denominator, separately, as in simple quantities.

17. Required the second root of

18. Required the square root of

SECTION IV

To extract the Square Root of a Number.

The following RULE for the extraction of the square root of a number, is derived directly from the method of extracting the square root of an algebraic quantity:

1. Separate the given number into periods of two figures each, beginning at the right hand. The period on the left may contain either one or two figures.

2. Find the greatest second power contained in the left-hand period, and write its root in the quotient's place.

3. Subtract the square of this root from the first period of the given quantity, and bring down the next two figures for a dividend.

4. Double the root already found, and put it in the divisor's place.

5. See how many times the divisor is contained in the new dividend, rejecting the right-hand figure; and annex the quotient both to the root already found ana to the divisor.

6. Multiply the divisor, thus increased, by the figure of the root last found, and subtract the product from the dividend.

7. Bring down the next two figures of the given number for a new dividend.

8. Double the whole root for a new divisor, and divide as before.

9. Proceed in this way, until the entire root of the given quantity is extracted.