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That is, when one power of a compound quantity is to be multiplied by any power of the same quantity, it may be expressed by adding the exponents, in the same manner as simple quantities.

The 2d power of (a + b)3 is (a + b)3 × (a + b)3

=

= (a + b) 3+3 = (a + b)3×2 = (a + b)°.

The 3d power of (2 a — d)* is

(2 a — d)11+1 = (2 a — d) 1×3 = (2 a — d)1.

That is, any quantity, which is already a power of a compound quantity, may be raised to any power by multiplying its exponent by the exponent of the power to which it is to be raised.

7. Express the 2d power of (3 b—c)*.

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10. Divide (7 m + 2 c)' by (7 m + 2 c)3.

If (a+b) is to be multiplied by any quantity c, it may be expressed thus: c (a+b)2. But in order to perform the operation, the 2d power of a + b must first be found.

c (n+b)2 = c
= c (a2 + 2 a b + b2) == a2 c + 2 a b c + b2 c

If the operation were performed previously, a very erroneous result would be obtained; for c (a+b) is very different from (ac+bc). The value of the latter expression is a2 c2 +2 ab c2 + b2 c2.

11. What is the value of 2 (a + 3b)3 developed as above? 12. What is the value of 3 b c (2 a — c)3 ?

13. What is the value of (a + 3 c2) (3 a—2b)*?

14. What is the value of (2 a-b)2 (a* + b c)*?

We have had occasion in the preceding pages to return from the second and third powers to their roots. We have shown how this can be done in numeral quantities; it remains to be shown how it may be effected in literal quantities. It is frequently necessary to find the roots of other powers as well as of the second and third.

The power of a literal quantity, we have just seen, is found by multiplying its exponent by the exponent of the power to which it is to be raised.

2

The second power of a3 is a3×3 = a; consequently the second root of a® is a

6

= a3.

The third power of am is a3m; hence the third root of a3

must be a

3m

=am.

The second root of a", then must be a.

m

a3

2 m
2

Proof. The second power of a is a = am.

In general, the root of a literal quantity may be found by dividing its exponent by the number expressing the root; that is, by dividing by 2 for the second root, by 3 for the third root, &c. This is the reverse of the method of finding powers.

It was shown above, that any power of a quantity consisting of several factors is the same as the product of the powers of the several factors. From this it follows, that any root of a quantity consisting of several factors is the same as the product of the roots of all the factors.

The third power of a b c3 is a b c°; the third root of a b3 c must therefore be a2 b c3.

Numeral coefficients are factors, and in finding powers they are raised to the power; consequently in finding roots, the root of the coefficient must be taken.

The 2nd root of 16 a b2 is 4 a2 b.

Proof. 4 ab × 4 a2 b = 16 aa b3.

When the exponent of a quantity is divisible by the number expressing the degree of the root, the root can be found exactly; but when it is not, the exponent of the root will be a fraction.

The second root of a is a3. The second root of a is a1.

The third root of a is a. The nth root of a is a

m

root of a" is an.

The nth

The root of a fraction is found by taking the root of its numerator and of its denominator.

This is evident from the me

thod of finding the powers of fractions.

The root of any quantity may be expressed by enclosing it in a parenthesis or drawing a vinculum over it, and writing a fractional exponent over it, expressive of the root. Thus

The 3d root of 8 a3 b is expressed

(8 a3 b)3 or 8 a3b3.

The root of a compound quantity may be expressed in the same way.

The 4th root of a +5 a b is expressed

(a2 + 5 a b)1 or ao + 5 a 6a.

When a compound quantity has an exponent, its root may be found in the same manner as that of a simple quantity.

The 3d root of (2 b — a)o is (2 b — a) * — (2 b — a)2.

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With regard to the signs of roots it may be observed, that all even roots must have the double sign; for since all even powers are necessarily positive, it is impossible to tell whether the power was derived from a positive or negative root, unless something in the conditions of the question shows it. An even root of a negative quantity is impossible. All odd roots will have the same sign as the power.

15. What is the second root of 9 a2 b*?

16. What is the third root of 125 a b c ?

17 What is the fifth root of 32 a1o xm r?

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20 What is the second root of (2 m — x)" ? 21 What is the 6th root of (3 a + x)m ?

XXXVII. Roots of Compound Quantities.

When a compound quantity is a perfect power, its root may be found; and when it is not a perfect power, its root may be found by approximation, by a method similar to that employed for finding the roots of numeral quantities.

First we may observe, that no quantity consisting of only two terms can be a complete power; for the second power of a binomial consists of three terms; that of a + x, for example, is a2 + 2 a x + x2. The quantity a+b is not a complete second power.

Let it be required to find the second root of

9 x* a® + 4 a2 b1 + 12 x2 a1 b2.

The root of this will consist of at least two terms. The second power of the binomial a + b is a2+2ab+b2. This shows that the quantity must be arranged according to the powers of some letter as in division, for the second power of either term of the root will produce the highest power of the letters in that term.

Arrange the above according to the powers of x.

9 x* a® + 12 x2 a* b2 + 4 a2 b1

The formula a2 + 2 a b + b2 shows that we should find the first term a of the root by taking the root of the first term; the same must be the case in the given example.

The root of 9 x1 a° is 3 x2 a3. Write this in the place of a quotient, and subtract its second power. Then multiply 3 x2 a3 by 2 for a divisor, answering to 2 a of the formula.

9 x* a2 + 12 x3 a′ b3 + 4 a3 b′ (3 x2 a3 + 2 a b2

9 x' a'

12 x* aa b2 + 4 a2 b′ (6 x2a2 +2 ab2
12 x' a' b'+4 a' bʻ

Divide the next term by the divisor. This gives 2 a b2 for the next term of the root. Raise the whole root then to the

second power and subtract it. Or, which is the same thing, since the second power of the first term has already been subtracted, write the quantity 2 a b' at the right of the divisor as well as in the root. Multiply the whole divisor as it then stands by the last term of the root. This produces the terms corresponding to 2ab+b2, = b (2a + b) of the formula. This produces 12x2 a b2 + 4 a2 b', which being subtracted, there is no remainder. Consequently the root is 3x a + 2 a b2 or 3x2 a3-2 a b2. The second power of both is the same. the double sign had been given to the first term of the root, the second would have had it also, and the positive and negative roots would have been obtained together.

Let it be required to find the 2d root of

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If

The process in this case is the same as in the last example. The second term of the root has the sign in consequence of the term 60 a b m2 of the dividend being affected with that sign. If the quantity had been arranged according to the powers of the letter b, thus, 25 b2. ·60 a b m2 + 36 a2 m2, the root would have been 5b- 6 a m2 instead of 6 a m2 5 b. Both roots are right, for the second powers of the two quantities are the

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