Separating two figures on the right of the result for decimals, we have 6,88; but 6,89 would be more exact, because the cube of this last number, although greater than 327, approaches it more nearly than that of 6,88.

If the proposed number contain decimals already, before we proceed to extract the root, we must place on the right as many ciphers, as will be necessary to render the number of decimal figures a multiple of 3. Let there be, for example, 0,07, we must write 0,070, or 70 thousandths, which gives for a root, 0,4. In order to arrive at a root exact to hundredths, we must annex three additional ciphers, which gives 0,070000. The root of the greatest cube contained in 70000 being 41, that of 0,07 becomes 0,41, to within a hundredth.

153. Hitherto I have employed the formula for binomial quantities only in the extraction of the square and cube roots of numbers; this formula leads to an analogous process for obtaining the root of any degree whatever. I shall proceed to explain this process, after offering some remarks upon the extraction of roots, the exponent of which is a divisible number.

We may find the fourth root by extracting the square root twice successively; for by taking first the square root of a fourth power, a4, for example, we obtain the square, or a2, the square root of which is a, or the quantity sought.

It is obvious also, that the eighth root may be obtained by extracting the square root three times successively, since the square root of a is a1, and that of a1 is a2, and lastly, that of a2

is a.

In the same manner it may be shown, that all roots of a degree, designated by any of the numbers 2, 4, 8, 16, 32, &c., that is, by any power of 2, are obtained by successively extracting the square

root.

Roots, the exponents of which are not prime numbers, may be reduced to others of a degree less elevated; the sixth root, for example, may be found by extracting the square and afterwards the cube root. Thus, if we take a and go through this process with it, we find by the first step a3, and by the second a; we may also take first the cube root, which gives a2, and afterwards the square root, whence we have a, as before.

154. I now proceed to treat of the general method, which I shall apply to roots of the fifth degree. The illustration will be

rendered more easy, if we take a particular example; and by comparing the different steps with the methods given for the extraction of the square and the cube root, we shall readily perceive, in what manner we are to proceed in finding roots of any degree whatever.

Let it be required then to extract the fifth root of 231554007. Now the least number, it may be observed, consisting of 2 figures, that is 10, has in its fifth power, which is 100000, six figures; we therefore conclude, that the fifth root of the number proposed contains at least two figures; this root may then be represented by a + b, a denoting the tens and b the units. The expression for the proposed number will then be

·(a + b)5 = a5 + 5 a1 b + 10 a3 l2 + &c.

I have not developed all the terms of this power, because it is sufficient, as will be seen immediately, that the composition of the first two be known.

2315,54007 | 47

1024

1291 5,4007 1280 therefore, 4 for the

Now it is evident, that as a3, or the fifth power of the tens of this root, can have no figure, that falls below hundreds of thousands, it does not enter into the last five figures on the right of the proposed number; we, therefore, separate these five figures. If there remained more than five figures on the left, we should repeat the same reasoning, and thus separate the proposed number into portions of five figures each, proceeding from the right to the left. The last of these portions on the left, will contain the fifth power of the units of the highest order found in the root. We find, by forming the fifth powers of numbers consisting of only one figure, that 2315 lies between the fifth power of 4, or 1024, and that of five, or 3125. We take, tens of the root sought; then subtracting the fifth power of this number, or 1024, from the first portion of the proposed number, we have for a remainder 1291. This remainder, together with the following portion, which is to be brought down, must contain 5 ab+10 a3 62+ &c. which is left, after a5 has been subtracted from (a+b)5; but among these terms, that of the highest degree is 5 ab, or five times the fourth power of the tens multiplied by the units, because it has no figure, which falls below tens of thousands. In order to consider this term by itself, we separate the last four figures on the right, which make no part of it, and the number 12915, remaining on the left, will contain this

term, together with the tens of thousands arising from the succeeding terms. It is obvious, therefore, that by dividing 12915 by 5 a1, or five times the fourth power of the four tens already found, we shall only approximate the units. The fourth power of 4 is 256; five times this gives 1280; if we divide 12915 by 1280, we find 10 for the quotient, but we cannot put more than 9 in the place of the root, and it is even necessary, before we adopt this, to try whether the whole root 49, which we thus obtain, will not give a fifth power greater than the proposed number. We find indeed by pursuing this course, that the number 49 must be diminished by two units, and that the actual root is 47, with a remainder 2209000; for the fifth power of 47 is 229345007; that is, the exact root of the proposed number falls between 47 and 48.

If there were another portion still, we should bring it down and annex it to the remainder, resulting from the subtraction of the fifth power found as above, from the first two portions, and proceed with this whole remainder, as we did with the preceding, and so on.

After what has been said, it will be easy to apply the rules, which have been given, as well in extracting the square and cube root of fractions, as in approximating the roots of imperfect powers of these degrees.

155. We may by processes, founded on the same principles, extract the roots of literal quantities. The following example will be sufficient to illustrate the method, which is to be employed, whatever be the degree of the root required.

We found in art. 143., the sixth power of 2 x3- 5 a3; we shall now extract the root of this power. The process is as fol

The quantity proposed being arranged with reference to the letter x, its first term must be the sixth power of the first term of the root arranged with reference to the same letter; taking then the sixth root of 64 x18, according to the rule given in art. 129., we have 2 x3 for the first term of the root required.

If we raise this result to the sixth power, and subtract it from the proposed quantity, the remainder must necessarily commence with the second term, produced by the developement of the sixth power of the first two terms of the root. But, in the expression (a + b)ε = a® + 6 a5 b + &c.

this second term is the product of six times the fifth power of the first term of the root by the second; and if we divide it by 6 a5, the quotient will be the second term b.

We must, therefore, take six times the fifth power of the first term 2 x3 of the root, which gives

6 x 32 x15 or

and divide, by this quantity, the term

192 x15,

960 a3 x15, which is the first term of the remainder, after the preceding operation; the quotient 5 a3 is the second term of the root. In order to verify it, we raise the binomial 2 x3 5 a3 to the sixth power, which we find is the proposed quantity itself.

If the quantity were such as to require another term in the root, we should proceed to find, after the manner above given, a second remainder, which would begin with six times the product of the fifth power of the first two terms of the root by the third, and which consequently being divided by 6 (2 x3-5 a3)5, the quotient would be this third term of the root; we should then verify it by taking the sixth power of the three terms. The same course might be pursued, whatever number of terms might remain to be found.

Of Equations with Two Terms.

156. EVERY equation, involving only one power of the unknown quantity combined with known quantities, may always be reduced to two terms, one of which is made up of all those which contain the unknown quantity, united in one expression, and the other comprehends all the known quantities collected together. This has been already shown with respect to equations of the second degree, art. 105., and may be easily proved concerning those of any degree whatever.

If we have, for example, the equation

a2 x5 — a5 b2 = b2 c3 + a c x5,

by bringing all the terms involving x into one member, we obtain a2 x5 — a cx3 = b4 c3 + a5 b2,

or

(a2 — a c) x5 = b2 c3 + n3 b2

Now if we represent the quantities

a2-ac by p, b4 c3 + a5 b2 by q,

the preceding equation becomes

p x3 = q ;

freeing 5 from the quantity, by which it is multiplied, we have

In general, every equation with two terms being reduced to the

taking the root then of the degree m of each member, we have

157. It must be observed, that if the exponent m is an odd number, the radical expression will have only one sign, which will be that of the original quantity (131).

When the exponent m is even, the radical expression will have the double sign; it will in this case be imaginary, if the quantity is negative, and the question will be absurd, like those of

P

which we have seen examples in equations of the second degree (131).