We find, in the second, the symbol of infinity, as in the preceding article, but the first presents this indeterminate form, o, of which we have already seen examples in articles 69 and 70; and before we pronounce upon its value, it is proper to examine, whether it does not belong to the case stated in art. 70; whether there is not some factor common to the numerator and denominator which the supposition of m = 1 renders equal to zero. It is here evident, that the numerator does not become 0 except by means of the factor m -m; we must therefore examine, whether this last has not some factor in common with the denominator 1-m. In order to avoid the inconvenience, arising from the use of the radical sign, let us make √m = n, then taking the squares, we have m=n2; the quantities therefore become but n - m n2, (1 — n) (1+n) (34); · n2 = n (1 — n), and 1 n2 = restoring to the place of n its value ✅m, we have √m — m = (1 —√m) √m a result the same, as that found in art. 119. a m = 1+ √m In the same manner we may reduce the second value of x, observing that * The example, which I have given at some length, corresponds with a problem resolved by Clairaut, in his Algebra, the enunciation of r It will be seen without difficulty, that we might have avoided radical expressions in the preceding calculations, by taking m2 to represent the ratio, which the squares of the two parts of the proposed number have to each other; m would then have been the square root, which may always be considered as known, when the square is known; but we could not have perceived from the beginning the object of such a change in a given term, of which algebraists often avail themselves, in order to render calculations more simple. It is recommended to the learner therefore, to go over the solution again, putting m2 in the place of m. Of the extraction of the square root of algebraic quantities. 121. We have sufficiently illustrated, by the preceding example, the manner of conducting the solution of literal questions. We have given also an instance of a transformation, namely, that of am into am, which is worthy of particular attention; since by means of it, we have been able to reduce the factors, contained under a radical sign, to the smallest number possible, and thus to simplify very much the extraction of the remaining part of the root. This transformation consists in taking the roots of all the factors which are squares, and writing them without the radical sign, as multipliers of the radical quantity, and retaining under the radical sign all those factors, which are not squares. This rule supposes, that the student is already able to determine, whether an algebraic quantity is a square, and is acquainted with the method of extracting the root of such a quantity. In order to this, it is necessary to distinguish simple quantities from polynomials. 122. It is evident, from the rule given for the exponents in which is as follows; To find on the line, which joins any two luminous bodies, the point where these two bodies shine with equal light. I have divested this problem of the physical circumstances, which are foreign to the object of this work, and which only divert the attention from the character of the algebraic expressions. These expressions are very remarkable in themselves, and for this reason I have developed them more fully, than they were done in the work referred to. multiplication, that the second power of any quantity has an exponent double that of this quantity. We have, for example, a1 xa1 = a2, a2 xa2 = a*, a3 xa3 = a, &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 Va2 = a1 or a, Vāa=a2, √ã3=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, c, in the expression 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 a6 b4 c2 = 8 as 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 a4 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 × 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, b2 and b, the first of which is a square; we have then b = 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 a4 b3 c2 = 36.2 a4 b2 .be1.& ; by collecting the factors, which are squares, it becomes 36 a4 b2 c4 × 2b c. Lastly, taking the root of the first product and indicating that of the second, we have 72a4 b3 cs6 a 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, ora2 +2ab+b2, contains the term +2 ab, which is not found in the expression a2 + b2. Let there be the trinomial 24 a b3 c +16 ac2 +9bo. 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 Now, whatever be the square root sought, if we suppose it ar ranged 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, 96, of the proposed quantity. The operation may be exhibited, as follows; 16 a4 c2 + 24 ao b3 c + 9 b* f 4 a2 c + 3 b3 root We begin by finding the square root of the first term, 16 a1 c2, and the result 4 a2 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. 4 We subtract from the proposed quantity, the square, 16 a1 c 2 of the first term, 4 a c, of the root; there remain then only the two terms 24 a2 b3 c + 9 bo. As the term 24 a b c is double the product of the first term of the root, 4 a2 c, by the second, we obtain this last, by dividing 24 a2 b3 c by 8 aa c, double of 4 a c, which is written below the root; the quotient 3 b3 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 9b, or rather, double of the first term of the root 8 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 a2 c, and multiply 8 ac+3bs by 3 63; 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 a2 c + 3 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 l2 + 2 1 p + p2, |