multiplication, that the second power of any quantity has an expo

nent double that of this quantity.

We have, for example,

a1 X a1 = a2, a3 × a2 = a1, a3 × a3 = ao, &c.

It follows then, that every factor, which is a square, must have an exponent which is an even quantity, and that the root of this factor is found by writing its letter with an exponent equal to half the original exponent.

Thus we have

a2 √ā1
= a1 or a, a = a2, √ a3 = a3, &c.

With respect to numerical factors, their roots are extracted, when they admit of any, by the rules already given.

Whence the factors a, b, c2, in the expression

64 a6 b4 c2,

are squares, and the number 64 is the square of 8; therefore, as the expression proposed is the product of factors, which are squares, it will have for a root the product of the roots of these several factors (121); and, consequently,

√64 a® b1 c2 = 8 a3 b2 c.

123. In other cases, different from the above, we must endeavour to resolve the proposed quantity, considered as a product, into two other products, one of which shall contain only such factors as are squares, and the other those factors which are not squares. To effect this, we must consider each of the quantities separately. Let there be, for example,

✔72 a1 b3 c5.

We see that among the divisors of 72, the following are perfect squares, namely, 4, 9, and 36; if we take the greatest, we have 72 36 X 2.

As the factor a is a square, we separate it from the others; passing then to the factor b3, which is not a square, since 3 is an odd number, we observe that this factor may be resolved into two others, a and b, the first of which is a square; we have then b2 = b2.b;

it is obvious also that

c5c4. c.

By proceeding in the same manner with every letter, whose exponent is an odd number, the quantity is resolved thus,

72 a b3 c5 36. 2 a1 b2. b c1.c;

by collecting the factors, which are squares, it becomes 36 a4 b2 c4 X 2bc.

Lastly, taking the root of the first product and indicating that of the second, we have

√72 a1 b3 c5 = 6 a2 b c2 √2bc.

See some examples of this kind of reduction, with the steps, by which they are performed;

It will be seen by the first of these examples, that the denominator of an algebraic fraction may be taken from under the radical sign by being made a complete square, in the same manner as we reduce the root of a numerical fraction (104.)

124. We now proceed to the extraction of the square root of polynomials. It must here be recollected, that no binomial is a perfect square, because every simple quantity raised to a square produces only a simple quantity, and the square of a binomial always contains three parts (34).

It would be a great mistake to suppose the binomial a + b to be the square root of a2 + b2, although taken separately, a is the root of a2, and b that of b2; for the square of a + b, or a2 + 2 a b + b2, contains the term + 2 ab, which is not found in the expression a2 + b2.

Let there be the trinomial

24 a2 b3 c + 16 a4 c2 + 9 bo.

In order to obtain from this expression the three parts, which
compose the square of a binomial, we must arrange it with
reference to one of its letters, the letter a, for example; it then
becomes
16 ac2 24 a2 b3 c + 9 bo®.

Now, whatever be the square root sought, if we suppose it arranged with reference to the same letter a, the square of its first term must necessarily form the first term, 16 a c2, of the proposed quantity; double the product of the first term of the root by the second must give the second term, 24 a2 b3 c, of the proposed quantity; and the square of the last term of the root must give exactly the last term, 9 bo, of the proposed quantity. The operation may be exhibited, as follows;

16 ac2 + 24 a2 b3 c + 9 bo § 4 a2 c + 3 b3 root
8a2c3b3

16 a4 c2

We begin by finding the square root of the first term, 16 a1 c2, and the result 4a2 c (122) is the first term of the root, which is to be written on the right, upon the same line with the quantity, whose root is to be extracted.

We subtract from the proposed quantity, the square, 16 a1 c2, of the first term, 4 a2 c, of the root; there remain then only the two terms 24 a2 b3 c + 9 bo.

As the term 24 a2 b3 c is double the product of the first term of the root, 4 a c, by the second, we obtain this last, by dividing 24 a2 b3 c by 8 a2 c, double of 4 a2 c, which is written below the root; the quotient 363 is the second term of the root.

The root is now determined; and, if it be exact, the square of the second term will be 966, or rather, double of the first term of the root 3 a2 c together with the second 3 b3, multiplied by the second, will reproduce the two last terms of the square (91); therefore we write + 363 by the side of 8 a c, and multiply 8 a2c+3b3 by 3 b3; after the product is subtracted from the two last terms of the quantity proposed, nothing remains; and we conclude, that this quantity is the square of 4 ac3 b3.

It is evident that the same reasoning and the same process may be applied to all quantities composed of three terms.

125. When the quantity, whose root is to be extracted, has more than three terms, it is no longer the square of a binomial; but if we suppose it the square of a trinomial, m + n + p, and represent by the sum m +n, this trinomial becoming now l+p, its square will be

12 + 2 1 p + p3,

in which the square la of the binomial m + n, produces, when developed, the terms m2 + 2mn+n2. Now, after we have arranged the proposed quantity, the first term will evidently be the square of the first term of the root, and the second will contain double the product of the first term of the root by the second of this root; we shall then obtain this last by dividing the second term of the proposed quantity by double the root of the first. Knowing then the two first terms of the root sought, we complete the square of these two terms, represented here by 12; subtracting this square from the proposed quantity, we have for a remainder

21p+p2,

a quantity, which contains double the product of l, or of the first binomial m+n, by the remainder of the root, plus the square of this remainder. It is evident, therefore, that we must proceed with this binomial as we have done with the first term m of the root.

Let there be, for example, the quantity

64a2 b c + 25 a2 b2 — 40 a3 b + 16 a1 + 64 b3 ca

80 a b2 c;

we arrange it with reference to the letter a, and make the same disposition of the several parts of the operation as in the above example.

16a1-40a3b+25a2b3—80ab2c+64b2c2 (4a2
+64a2bc

5ab+8bc

8a2

-16a

5ab 8a210ab8bc

1st rem.-40a3b+25a3b2-80ab2c+64b2 ca

We extract the square root of the first term 16 a1, and obtain 4a2 for the first term of the root sought, the square of which is to be subtracted from the proposed quantity.

We double the first term of the root, and write the result, 8 a2, under the root; dividing by this the term 40 a3 b, which begins the first remainder, we have 5 a b for the second term of the root; this is to be placed by the side of 8 a2, we then multiply the whole by this second term, and subtract the result from the remainder, upon which we are employed.

Thus we have subtracted from the proposed quantity the square of the binomial 4 a2-5 ab; the second remainder can contain only double the product of this binomial, by the third term of the root, together with the square of this term; we take then double the quantity 4 a2-5 ab, or

8a210 ab,

which is written under 8 a5 a b, and constitutes the divisor to be used with the second remainder; the first term of the quotient, which is 8 b c, is the third of the root.

This term we write by the side of 8 a2 — 10 ab, and multiply the whole expression by it; the product being subtracted from the remainder under consideration, nothing is left; the quantity proposed, therefore, is the square of

The above operation, which is perfectly analogous to that, which has been already applied to numbers, may be extended to any length we please.

Of the formation of Powers and the extraction of their Roots.

126. THE arithmetical operation, upon which the resolution of equations of the second degree depends, and by which we ascend from the square of a quantity to the quantity, from which it is derived, or to the square root, is only a particular case of a more general problem, namely, to find a number, any power of which is known. The investigation of this problem leads to a result, that is still termed a root, the different kinds being called degrees, but the process is to be understood only by a careful examination of the steps by which a power is obtained, one operation being the reverse of the other, as we observe with respect to division and multiplication, with which it will soon be perceived that this subject has other relations.

It is by multiplication, that we arrive at the powers of entire numbers (24), and it is evident, that those of fractions also are formed by raising the numerator and denominator to the power proposed (96).

So also the root of a fraction, of whatever degree, is obtained by taking the corresponding root of the numerator and that of the denominator.

As algebraic symbols are of great use in expressing every thing, which relates to the composition and decomposition of