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XXXVIII.

Extraction of the Roots of Compound Quantities of any Degrée.

By examining the several powers of a binomial, and observing that the principle may be extended to roots consisting of more than two terms, we may derive a general rule for extracting roots of any degree whatever.

(a + x)' = a + x a + x

a2+ a x

ax + x2

(a + x)2= a2 + 2 a x + x2

a + x

a3 + 2 a3 x + a x2

a2 x + 2 a x2+203

(a + x)3 = a3 +3 a2 x + 3 a x2+003

a + x

a* + 3 a3 x + 3 a2 x2 + a x3

a3x+3a2x2 + 3 ax3 + x*

(a + x)' = a* + 4 a3 x + 6 a2 x2 +4 α x3 +x*

a + x

ao + 4 a*x + 6 a3 x2 +4 a3 x3 + a xa

a*x +4 a2x2 + 6 a2x2 + 4 a x + 20°

· (a + x)3 = a* + 5 a* x + 10 a3 x2 + 10 a3 x3 + 5 a x + x3

By examining these powers, we find that the first term is the first term of the binomial, raised to the power to which the binomial is raised. The second term consists of the first term of the binomial one degree lower than in the first term, multi

plied by the number expressing the power of the binomial, and also by the second term of the binomial. This will hereafter be shown to be true in all cases.

The application will be most easily understood by a particular example.

Let it be required to extract the 5th root of the quantity

32 a1o 80 a3 b2+ 30 ao bo— 40 a* b3 + 10 a2 b12 — b13 (2 a2 — b3 32 a1

*

Dividend.
80 as b3

80 a divisor.

The quantity being arranged according to the powers of a, I seek the fifth root of the first term 32 a'. It is 2 a2. This I write in the place of the quotient in division. I subtract the fifth power of 2 a2, which is 32 a', from the whole quantity. The remainder is

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The second term of the fifth power of the binomial a +x being 5 ax shows that if the second term in this case be divided by five times the 4th power of 2 a2, the quotient will be the next term of the root. The 4th power of 2 a2 is 16 a and 5 times this 80 as. Now Now S0 a 3 being divided by 80 a gives 63 for the next term of the root. Raising 2 a2 to the fifth power, it produces the quantity given. If the root contained more than two terms it would be necessary to subtract the 5th power of 2 a2 — b3 from the whole quantity; and then to find the next term of the root, divide the first term of the remainder by five times the 4th power of 2 a2 - b3. 'The first term only however would be used which would be the same divisor that was used the first time.

When the number expressing the root has divisors, the roots may be found more easily than to extract them directly. The second root of a1 is a2, the second root of which is a. Hence the 4th root may be found by two extractions of the second The second root of a is a3, or the 3d root of a® is a2. Hence the 6th root may be found by extracting the 2d and 3d roots. The 8th root is found by three extractions of the 2d root, &c.

root.

Examples.

1. What is the 3d root of

6x+x40x396x-64?
x3

2. What is the third root of

15 x — 6 x + x3 — 6 x3 — 20 x3 + 15 x2 + 1 ?

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3. What is the 4th root of

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216 a2x2-216 a x + 81 x + 16 a*-96 a* x?

4. What is the 5th root of

80x40x+32x80x-1+10x?

XXXIX. Extraction of the Roots of Numeral Quantities of any Degree.

By the above expression of the several powers, we may extract any root of a numeral quantity. Let us take a particular example.

What is the 5th root of 5,443,532,400,000?

In the first place we observe that the 5th power of 10 is 100000, and the 5th power of 100 is 10000000000. Therefore if the root contains a figure in the ten's place, it must be sought among the figures at the left of the first five places counting from the right. Also if the root contains a figure in the hundred's place, it must be sought at the left of the first ten figures. This shows that the number may be divided into periods of five figures each, beginning at the right. The number so prepared will stand

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In the first place I find the greatest 5th power in 544. It is 243, the root of which is 3. I write 3 in the root, and subtract 243, the 5th power of 3, from 544. The remainder must contain 5 a x + 10 a3 x2+, &c. The 3, that part of the root already found, and which, by the number of periods, must be 300, answers to a in the formula. 5 a*, that is, five times the fourth power of 300 will form only an approximate divisor, since the remainder consists of several terms besides 5 a* x; still it will enable us to judge very nearly, and we shall find the right number after one or two trials. As the fourth power of 30 will have no significant figure below 10000, (we may consider 3 to be in the ten's place, with regard to the next figure to be found,) we may bring down only one figure of the next period to the remainder for the dividend, and use 5 times the fourth power of 3 for the divisor. The dividend is 3013 and the divisor 405. The dividend contains the divisor at least 6 times, but probably 6 is too large for the root. Try 5. This gives for the first two figures 35. Raise 35 to the 5th power and see if it is equal to 544,35324. It will exceed it. Therefore try 4. The fifth power of 34 is 544,35324. Hence 34 is right. Subtract this from the number, there is no remainder. There is still another period, but it contains no significant figure, therefore the next figure is 0, and the root is 340. The 5th power of 340 is 5,443,532,400,000. If there had been a remainder after subtracting the 5th power of 34, it would have been necessary to bring down the next figure of the number to it to form a dividend, and then to divide it by 5 times the 4th power of 34; and to proceed in all respects as before.

The process of extracting roots above the second is very tedious. A method of doing it by logarithms will hereafter be shown, by which it may be much more expeditiously performed.

Examples.

1. What is the 5th root of 15937022465957?

2. What is the 4th root of 36469158961 ?

For this, the fourth root may be extracted directly, or it may be done by two extractions of the second root. Let the learner do it both ways.

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It is very important to remember how these quantities may be separated into factors. Since multiplication is performed by adding the exponents, and division by subtracting them, any quantity may be separated into as many factors as we please, by separating the exponent into parts. Thus,

a3 = a3 × a2 = a × a1 = a × a2 X a2

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The sum of all the exponents in the last expression is 5. Logarithms are of the same nature as these exponents, and afford as great a facility in operating upon numbers, as these do upon letters. And the operations are performed in the same way, as will be explained hereafter.

If the learner should ever have occasion to read other treatises on mathematics, he will generally find the roots expressed by what are called radical signs. The second root is expressed with the sign√, the third root the same sign with the index of the root over it. The 4th root is, &c.

= √ a

3

Va

3

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They will be easily understood if the radical sign be removed, and the exponents divided by the index of the root or the quantity enclosed in a parenthesis, and the root written over it. The expression 5 a 63 becomes

4

5a a‡ b# = (5 a2 b3)‡.

The expression √ u2 + 6a is equivalent to (a2 + b2)$

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