51. By means of the binomial theorem, a trinomial, or even a quadrinomial quantity, may be raised to any power. Let it be required, for example, to find the fourth power of a+b+c. By considering a+b as one quantity, we shall have, (a+b+c)1=(a+b)+c)* = (a+b)1+4(a+b)3c + 6(a+b)2c2+ 4(a+b)c+c, as before. Now, expanding these terms by the theorem, we shall have a+b+c)a = a*+4a3b+6a2b2+4ab3+b*+ 4a3c+12a2bc+12ab2c+4b3c+6a2c2+12abc2+6b2c2+4ac3+4bc3+ c. In the same manner, (a-b+c—d)3—(a—b)+(c—d)3 — (a—b)3+3(a—b)2 (c—d)+3(a—b) (c—d)2 + (c—d)3—a3-3a2b 3ab2b+3a2c-3a2d6abc+6abd+3b2c-3b2d3ac2—3bc2— 6acd+6bcd+3ad2—3bd2+c3—3c2d+3cd2-d3. EXAMPLES. Ex. 1. Required the third power of (a-x). On the Evolution of Algebraic Quantities. 52. Evolution is the reverse of involution, and therefore explains that process by which, when the power is given, the root may be found. It has been already observed, that, when a quantity is multiplied once by itself, the product is called the square of the given quantity; we may now observe farther, that the quantity itself, in reference to this square, is denominated the square root. In the same manner, if a quantity be multiplied twice by itself, the product is called the cube of the given quantity, which, in reference to this cube, is denominated the cube root. The term biquadrate root, is sometimes, but seldom used. Higher roots than the third or cube, are generally designated by the corresponding ordinal number, as the fourth, fifth, &c. root. 53. From this, it appears, that to find the square root of a quantity, is nothing else, than to find such a quantity as, when multiplied once by itself, will produce the given quantity. Thus, 64=8×8, therefore, the square root of 64 or √√648; a2 = axa, and, therefore, a is the square root of a2 or a2a, where the root may be either positive or negative; because +x+ gives +, and -x- gives also +. In the same manner, the finding of the cube root consists in finding such a quantity as when multiplied twice by itself will produce the given quantity. Thus, 27=3×3×3, therefore, the cube root of 27 or √273; a3 = axaxa, and, therefore, Va3 =a. 54. The whole statements now made regarding the extraction of roots, may be presented in a more general and condensed form, by saying, that the finding of any root of a quantity, consists in finding that factor, which, when multiplied into itself, once less than the units in the index of the root, will produce the given quantity. Thus, the fifth root of 32 or √32=2, since 32=2×2×2×2×2; the mth root of am or m/am=a. The cube root of -3 or3 ——x, because -X. The root of a fraction is found by extracting both the root of its numerator and de -X, The root of a 27t of a mixed number or quantity, will be found, by first reducing it to an improper fraction, and then extracting; thus, 55. If the quantity under the radical sign does not admit of having the required root extracted, it is called a surd; thus, √5, √9, Va2, are surd quantities. The application of the fundamental rules to the management of surds, will be shown afterwards. 56. Any even root of a negative quantity is unassignable. Thus, -a, -a, &c. cannot be determined, because all the even powers of +a ora are positive. Quantities, therefore, of the form 4, -a, &c. have no real root, √ and are, therefore, called impossible. 57. A general rule for the extraction of the square root of a compound quantity may be readily obtained, by observing how the terms of the root are to be derived from those of the power. The square of a+b, according to the table (p. 82.) is a2+2ab+b2, which consists of the squares of the two terms a2 and b2, together with twice their product 2ab. If, therefore, when the terms are ranged according to the powers of a, we extract the square root of the first term (a2), we will obtain the first term of the root (a); and if the next term (2ab) be divided by the double of this part of the root now found, that is, by 2a, this division will give b the second term of the root. Hence the process will stand as follows: a2+2ab+b2(a+b a2 2a+b)2ab+b2 The square root of the first term a2, is a, which, being placed in the root, and its square (a2) subtracted from the corresponding term of the power (a2), leaves no remainder. The next two terms 2ab and b2 are brought down, the first of which, 2ab, being divided by 2a, gives b the second term of the root; and as 2ab+b2 = (2a+b)b, if to 2a the term b be added, and this sum be multiplied by b, the product is 2ab+b2, which, when subtracted, leaves no remainder. The square of a+b+c, is (No. 51.) = a2+2ab+b2+2ac+ 2bc+c2, where the root may be derived by a mere continuation of the process, explained in the preceding example. a2+2ab+b2+2ac+2bc+c2(a+b+c a2 2a+b)2ab+b2 2a+2b+c)2ac+2bc+c2 Thus, the two first terms a+b of the root being found as before, the remaining three terms 2ac+2bc+c2 of the power are brought down. The part of the root a+b, already found, is doubled, which gives 2a+2b; the first term of the remainder 2ac being divided by 2a gives c, which, being added to both the divisor and root, and the whole of the former multiplied by it, gives 2ac+2bc+c2, and this being subtracted leaves no remainder. In this manner, the following examples are solved: 58. By the same process, we are enabled to discover the square roots of numbers. Let 345 be selected as a root; 345=300+40+5, now the square of this may be found as that of a+b+c, and will be 90000+24000+4600+400+25, let the square root of this be extracted. 90000+24000+4600+400+25(300+40+5 90000 600+40)24000+4600+400 600+80+5)3000+400+25 3000+400+25 Now, it is manifest that the result will not be altered by collecting those numbers into one sum, which stand in the same line, and leaving out the ciphers which merely serve to indicate the rank of the digits; let this be done, and the above process will assume the following condensed form, from which the following rule is evolved: 119025(345 9 64)290 256 685)3425 Divide the number whose root is to be extracted, into periods of two figures each, beginning from the unit's place. Find the greatest number, whose square is contained in the first period, and this will be the first figure of the root. Subtract the square of this figure from the first period, and to the remainder annex the next. Divide this number (omitting the last figure) by twice that part of the root already found, annex the result to the root and also to the divisor; then subtract the product of the divisor thus augmented, by that figure of the root last obtained; and to the remainder annex the next period, with which proceed as before; and so on till the given number be exhausted. The reason of dividing the number, whose root is to be extracted, into periods of two figures each, is because the square of a number can never have more than twice as many places as its root, and at the least but one less; consequently there will always be just as many figures in the root as there are pairs of figures in the given number, the left hand part consisting in some cases of a half period or one figure. Ex. 2. Required the square root of 7645225. 7645225(2765 4 47)364 546)3552 5525)27625 |