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same. The second power a b is the same as that of b One is the positive and the other the negative root. If the double sign be given to the first term of the root, both results will be produced at the same time in either arrangement.

the

25 b3 — 60 a b m2 +36 a2 m2 (±5b6a m2

2562

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* -60 a b m2 + 36 a2 m2 (±10b 6am2
·60 ab m2 +36 a2 m*

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to, and the to +. the sign. The first value is 6 am2 - 5 b.

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When the quantity whose second root is to be found, consists of more than three terms, it is not the second power of a binomial, but of a quantity consisting of more than two terms. Suppose the root to consist of the three terms m+n+p. If we represent the two first terms m+n by l, the expression becomes l+p, the second power of which is

l2 + 21p+p2.

Developing the second power l2 of the binomial m +n, it becomes m2 + 2 m n + n2. This shows that when the quantity is arranged according to the powers of some letter, the second root of the first term will be the first term m of the root. If m2 be subtracted, and the next term be divided by 2 m, the next term n of the root will be obtained. If the second power of m + n or l2 be subtracted, the remainder will be 2 1 p + p3. If the next term 21 p be divided by 21 equal to twice m+n, the quotient will be P, the third term of the root. The same principle will extend to any number of terms.

It is required to find the second root of

4 a* + 12 a3 x + 13 a2 x2 + 6 α x3 + x3.

Let this be disposed according to the powers of a or of x.

x+6α x3+ 13 a2x2+ 12 a3 x + 4 a1 (x2 + 3 ax + 2 a root.

*

1st dividend.

6 a x3 + 13 a2x2

(2x2+3ax 1st divisor.

6 α x3 + 9 a2x2

2d divid. * 4a2x2 + 12 a3 x + 4 a1 (2x2 + 6 a x + 2 a2 2d. di. 4 a x + 12 a3 x + 4 a'

*

The process is so similar to that of numeral quantities that it needs no farther explanation.

The double sign need not be given to the terms during the operation. All the signs may be changed when the work is done, if the other root is wanted. This will seldom be the case when all the terms are positive; but when some of the terms are negative, if it is not known which quantities are the largest, the negative root is as likely to be found first as the positive. When this happens the positive will be found by changing all the signs.

Examples.

1. What is the second root of

4a3 x + 6 a2x2 + a + x + 4 α x3?

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3. What is the second root of

-420 +420 + 12x36x + x2 + 9?
4x+4x+12.x3.

4. What is the second root of

x+20x3+25x+16+4x+10x+24 x?

XXXVIII.

Extraction of the Roots of Compound Quantities any Degree.

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.

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

a + x

a3 +4 a* x + 6 a3 x2 +4 α2 x3 + a xc4

a*x + 4 ́a3 x2 + 6 α3 x3 + 4 α x2 + x3

(a + x)3 = a3 + 5 a* x + 10 a3 x2 + 10 a2 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 as b3+ 80 a® b®— 40 a* bo + 10 a2 b12 — b1 (2 a2 — b3

32 al

*

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 a1o. 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 a1o, from the whole quantity. The remainder is

· 80 a3 b3 + 80 a® bo &c.

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The second term of the fifth power of the binomial a + x being 5 a* x 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 a* is 16 a3 and 5 times this 80 a. Now 80 a b3 being divided by 80 as gives b3 for the next term of the root. Raising 2 a2 — b3 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 a is a2, the second root of which is a. Hence the 4th root may be found by two extractions of the second root. 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.

Examples.

1. What is the 3d root of

6 x3 + x3 —40x3 + 96 x — 64 ?

2. What is the third root of

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

3. What is the 4th root of

216 a2x2-216 α x3 + 81 x + 16 a* —96 a3 x?

4. What is the 5th root of

80x40x2+ 32 x3 — 80 x1 — 1+10x?

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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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