3

x2 + 6 a x2 + 13 a2x2 + 12 a3 x + 4 a1 (x2 + 3 a x + 2 a2 root.

1st dividend.

6 a x3 + 13 a2x2 (2x2 + 3 ax 1st divisor.

6 a x3 + 9 a2x2

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

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+6a2x2 + a* + x + 4 ax3?

3. What is the second root of

-420 +420 + 12 x3 — 6 x + x2 + 9?

4. What is the second root of

x+20x3 +25 x2 + 16 +4x+10x + 24 x?

XXXVIII.

Extraction of the Roots of Compound Quantities of 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.

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

a + x

a

a3 + 4 a* x + 6 a3 x2 +4 a2 x3 + α x*

a*x + 4 a3x2 + 6 a2 x3 + 4 a x2 + 205

(a + x)3 = a3 + 5 a* x + 10 a3 x2 + 10 a2 x3 + 5 α x1 + 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— 30 a3 b3+ 30 ao bo— 40 a* bo + 10 a2 b12 — b13 (2 a2 — b3 32 a'

The quantity being arranged according to the powers of a, I seek the fifth root of the first term 32 a1. 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

80 a3 b3 + 80 a® bo &c.

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 a2 is 16 ao and 5 times this 80 as. Now Now 80 a b3 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 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 — 40 x3 + 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 a x3 +81 x + 16 a1-96 a3 x?

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

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 244. 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 a1, 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 544,25324. 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.