the sign of every even root of a positive quantity, and the sign of every odd root of any quantity the same as that of the quantity.

If the given quantity is a fraction, it follows, from Art. 204, that we may take the square root of both its terms for the required root.

5. Find the square root of 9 a1 b2 c-12.

Ans. 3a2bc-6.

6. Find the square root of 625 a12 c.

Ans. ±25 ac2.

Here, as we can not extract the root of c, its root is indicated in the result by the fractional exponent.

7. Find the mth root of a b-mn.

8. Find the cube root of · 8a-3 b6 x-2.

1

Ans. ab-n.

10. Required the value of (3 am b2n d2)n.

12. Required the square root of (a + x)2 b2 x y1.

- y)".

Ans. ±(a+x) b x2 y2.

SQUARE ROOT OF POLYNOMIALS.

224. The manner of forming the square of a polynomial must, by reversing the process, lead to the discovery of its root. If we take any binomial, as a+b, we have

(a + b)2 = a2+2ab+b2;

and the last two terms of this expression factored give

(2a + b) b.

1. Let us now reverse the involution, and discover how the root ab may be derived from the square a2+2ab+b2.

(2 a + b) b, for a remainder or dividend. Dividing the first term of the dividend, 2ab, by 2a, which is double the first term of the root, we obtain b, the other term of the root, which, connected to 2a, completes the divisor, 2ab. Multiplying this divisor by the last term of the root, b, and subtracting the product, 2 a b + b2, from the dividend, we have no remainder.

By a like process, a root consisting of more than two

terms may be found from its square, since all such roots can be expressed in a binomial form. Thus,

and its square,

a+b+c= (a + b) + c,

a2+2ab+b2+2ac+2bc+c2=(a+b)2+2(a+b)c+c2,

which, factored, gives

(a+b+c)2 = a2 + (2 a + b) b + (2a + 2 b+c) c.

Hence the following

RULE.

Arrange the terms according to the powers of some letter.

Find the square root of the first term, write it as the first term of the root, and subtract its square from the given polynomial.

Divide the first term of the remainder by double the part of the root already found, and annex the result to the root, and also to the divisor.

Multiply the divisor as it now stands by the term of the root last obtained, and subtract the product from the dividend.

If there are other terms remaining, continue the operation in the same manner as before.

Since all even roots have the double sign ± (Art. 223), the square root obtained by the rule will remain a root when all the signs are changed. In fact, in the example above, the first term in the root might have been found by taking ―a as the root of a2, and continuing the operation in the same manner, we should have α b for the

Here, it will be observed that the successive subtrahends in the operation, 9 x1, (6 x2 - 2 x) (-2x), and (6 x2-4x+2) 2, in the aggregate equal (3− 2 x + 2)2, in accordance with the formula upon which the rule is based.

3. Find the square root of 4x4 — 4x3-3x2+2x+1.

4. Find the square root of 4 at 16 a3 24 a2-16 a +4.

6. Extract the square root of (x+x-1)2 - 4 (x — x−1).

225. No rational binomial is an exact square; but, by

the rule, the approximate root may be found.

7. Find the approximate square root of 1+x to three

8. Find the approximate square root of a + b to three

10. Find the approximate square root of a2+x2 to five terms.

Ans. a+

5x8

+

+ etc.

11226. When a trinomial is a complete square, two of the terms are squares, and the other term is twice the product of their square roots (Art. 90); hence, if such a trinomial be arranged according to the powers of either of its letters,

The square roots of its extreme terms, united by the sign of its middle term, will be the square root of the trinomial.

11. Find the square root of 9 y1 — 6 x3 y2 + x6.

227.

CUBE ROOT OF POLYNOMIALS.

An investigation of the formation of a polynomial cube, by reversing the process, must lead to the discovery of its root. If we take any binomial, as a + b, we have

(a + b)3 = a3 + 3 a2 b + 3 a b2 + b3,

and the last three terms of this expression, factored, give (3 a2 + 3 a b + b2) b.