23. Root of a Quotient.

(v.) The rth root of a quotient is equal to the quotient obtained by dividing the rth root of the dividend by the rth root of the divisor.

(The following proof may be omitted when the chapter is read for the first time.)

If a and b be both positive, b being different from zero, and we represent the value of the positive fraction a/b by ƒ and of the whole number b by w, we have, applying

(i) above

vb.

MENTAL EXERCISE XVIII. 3

Simplify each of the following roots of quotients:

SQUARE ROOTS OF POLYNOMIALS

24. We have shown in Chap. XII. §§ 18, 19, a method for obtaining by inspection the equal factors, and hence the square root, of a trinomial which is a square.

We will now present another method.

Representing any trinomial square by a2 + 2 ab + b2, we may obtain the square root as follows:

Arranged according to descending powers of a, we may write

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

(1) We may obtain the first term in the required square root by taking the square root, a, of the first term, a2.

Subtracting the square of a from the given trinomial, we obtain as a first remainder 2 ab + b2, which may be written in the form (2a + b)b.

The second term of the required square root may be found by dividing the first term 2 ab of the first remainder 2 ab + b2, arranged according to descending powers of a, by a trial divisor 2 a, which is formed by multiplying by 2 the part of the root already found.

From (1) it appears that by increasing 2a by b we may obtain & complete divisor, 2a + b, which when multiplied by b will produce the terms 2 ab + b2 of the first remainder.

Subtracting the product of 2 a + b and b, that is, 2 ab + b2, from the first remainder, 2 ab + b2, we obtain zero as a second remainder. The steps of the process are shown below:

Ex. 1. Find the square root of 9xa — 42 x2y3 + 49 y®.

25. We will now show that the steps of the process may be repeated to obtain the square root of any polynomial square which contains more than three terms.

The square of a polynomial may be written as follows:

For four terms, (a + b + c + d)2 = a2 + 2 ab + 2 ac + 2 ad)

Factoring the groups of terms which appear in the vertical columns, the identities (1), (2), and (3) become respectively:

(1)

(2)

(a + b)2= a2 + (2a + b)b. (a+b+c)2 = a2 + (2a + b)b + (2a + 2 b+c)c. (a+b+c+d)2= a2 + (2a + b)b + (2a + 2b+c)c + (2a +2b+2c+ d)d. (3)

It may be seen that, with each new letter added on the left, a new group of terms in parentheses is added on the right. This new group consists of the product of twice the sum of all previous letters plus the last letter, multiplied by the last letter.

In (1), (2), and (3) the expressions in parentheses are in each case the complete divisors used in the extraction of the square root of a polynomial at the successive stages of the process.

Ex. 2. Find the square root of

a2+2ab+b2 + 2 ac + 2 bc + c2 + 2 ad + 2 bd + 2 cd + d2.

√a2= a. a2+2ab+b2+2 ac+2 be+c2+2 ad+2 bd+2 cd+d2 | a+b+c+d

26. The method employed in §§ 24 and 25 for extracting the square root of a polynomial may be stated in the following form as a rule.

Rule for finding the principal square root of a polynomial square.

Write the given polynomial according to descending or ascending powers of some letter of arrangement.

Extract the square root of the first term and write the result as the first term of the required square root.

Subtract the square of the first term of the root from the given polynomial, and arrange the first remainder according to the powers of the letter of arrangement, and in the same order as before.

Divide the first term of the remainder by twice the first term of the root, write the quotient as the second term of the root, and add it also to the trial divisor to form the complete divisor.

Subtract from the first remainder the product of the complete divisor multiplied by the term of the root last found, and arrange the remainder, if there be one, as a second remainder.

Repeat the process, using as a trial divisor at each stage of the work twice the part of the root already found,

Ex. 3. Find the square root of 29 a2 — 40a5+ 16ao — 46 a3+ 4 + 49 aa— 12a. Arranged according to descending powers of a we have

√16a6 = 4a3. | 16a6 — 40a5+49a* — 46a3 +29a2 — 12a+4 | 4a3-5a2+3a−2 (4 a3)2 = 16a6

In practice the student should obtain the successive terms of the root by performing the divisions, — 40 a5 ÷ 9 a3 = − 5 a2, etc., mentally.

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EXERCISE XVIII. 4

Find the square roots of the following expressions :

1. a1+2a3 — a2 2a + 1.

2.

-

16.xt 24x3 + 25 x2

3. 81 x

54 x3 + 81 x2 + 24 x + 16.

4. 9x624x5 + 22 x1 + 38 x3 + 41 x2 + 10x + 25.

5. 256-30 b* ·20b3 +962 + 12b + 4.

6. 36 a2 + 48 ab + 12 ac + 16 b2 + 8 bc + c2.

27. A process for finding the cube root of a polynomial which is

a cube may be developed as follows:

We know that (a + b)3 = a3 +- 3 a2l + 3 ab2 + l3.