a3, &c. A simple quantity is raised to any power, by multiplying the index of every factor in the quantity by the exponent of the power, and prefixing the proper sign determined by the last article. Thus, am raised to the nth power is am, Because am X am × am.....to n factors, by the rule of multiplication, is amn; also, a bn=a bab×a b×&c. to n factors, or axa+a....to n factors × bxbxb...to n factors anxon; and a2 b3 c raised to the fifth power is a b15 c5. Also, — am raised to the nth power is amn; where the positive or negative sign is to be prefixed, according as n is an even or odd number. If the quantity to be involved be a fraction, both the numerator and denominator must be raised to the proposed power. If the quantity proposed be a compound one, the involution may either be represented by the proper index, or it may actually take place. Let a+b be the quantity to be raised to any power. a+b a+b a2+a b +a b+b2 axbor a+2ab+b the sq. or 24 power a+b a3+2 æ2 b+a b3 a+03 or a3+3 a2 b+3 a b2+b3 the 3d pr. a+b a++3 a3 b+3a2 b2+ a b3 + a3 b+3a2 b2+3 a b3+b4 a+b)4 or a++3 a3 b+6a2 b2+4 a b3+64 the fourth power. If b be negative, or the quantity to be involved be a-b, wherever an odd power of b enters, the sign of the term must be negative. - Hence, a − b ) + = a+ — 4 a3 b+6 a2 b2 ·4 a 63+6. EVOLUTION, or the extraction of roots, is the method of determining a quantity, which, raised to a proposed power, will produce a given quantity. Since the nth power of am is amn, the ath root of amn must be am; i. e. to extract any root of a single quantity, we must divide the index of that quantity by the index of the root required. When the index of the quantity is not exactly divisible by the number which expresses the root to be extracted, that root must be represented according to the notation already pointed out. Thus the square, cube, fourth, nth root of a+x, are respectively represented by (a2 + x2)1‚ (a2 + x2)‡‚ (a) 1 (a2 + x)n; the same roots of (a2+x2)—1,arerepresentedby (a+) (a2+x3) ̃ ̄ ̃3‚ (a2+x3) ̄ ̄2, (a2+x2) 7. or If the root to be extracted be expressed by an odd number, the sign of the root will be the same with the sign of the proposed quantity. If the root to be extracted be expressed by an even number, and the quantity proposed be positive, the root may be either positive or negative. Because either a positive or negative quantity, raised to such a power, is positive. If the root proposed to be extracted be expressed by an even number, and the sign of the proposed quantity be negative, the root cannot be extracted; because no quantity, raised to an even power, can produce a negative result. Such roots are called impossible. Any root of a product may be found by taking that root of each factor, and multiplying the roots, so taken, together. Thus, (a b)nan X bn; because each of these quantities, raised to the nth pow. er, is a b. 2 In a=b, then an X an=an; and in the 8 pts a2+2ab+b2 (a+b a1 2 a+b)2 a b+b2 2 a b+b2 Since the square root of a2+2 a b+b2 is a+b, whatever be the values of a and b, we may obtain a general rule for the extraction of the square root, by observing in what manner a and b may be derived from a2+2 a b+b2. Having arranged the terms according to the dimensions of one letter, a, the square root of the first term a' is a, the first factor in the root: subtract its square from the whole quantity, and bring down the remainder 2 a b+b; divide 2 a b by 2 a, and the result is b, the other factor in the root; then multiply the sum of twice the first factor and the second (2a-+6), by the second (b), and subtract this product (2 a b+6) from the remainder. If there be no more terms, consider a+b as a new value of a; and the square, that is a'+2 a b+b2, having, by the first part of the process, been subtracted from the proposed quantity, divide the remainder by the double of this new value of a, for a new factor in the root; and for a new subtrahend, multiply this factor by twice the sum of the former factors increased by this factor. The process must be repeated till the root, or the necessary approximation to the root, is obtained. Ex. 1. To extract the square root of a2+2ab+b2+2 a c+2 b c+c2. a+2ab+b+2 ac+2bc+c2 (a+b+c is The cube root of a3+3 a2 b+3 a baxb3 is a+b; and to obtain a+b from this compound quantity, arrange the terms as before, and the cube root of the first term, a3, is a, the first factor in the root: subtract its cube from the whole quantity, and divide the first term of the remainder by 3 a2, the result is 6, the second factor in the root: then subtract 3 a2 b+3 a b2+b3 from the remainder, and the whole cube of a+b has been subtracted. If any quantity be left, proceed with a+b as a new a, and divide the last remainder by 3.a for a third factor in the root; and thus any number of factors may be obtained. ON SIMPLE EQUATIONS. If one quantity be equal to another, or to nothing, and this equality be expressed algebraically, it constitutes an equation. Thus, a= b-x is an equation, of which a forms one side, and 6-x the other. When an equation is cleared of fractions and surds, if it contain the first power only of an unknown quantity, it is call ed a simple equation, or an equation of one To find the value of an unknown quantity in a simple equation. Let the equation first be cleared of fractions, then transpose all the terms which involve the unknown quantity to one side of the equation, and the known quantities to the other; divide both sides by the coefficient, or sum of the co-efficients, of 19 x=95 x= 95 =5. 19 If there be two independent simple equations involving two unknown quantities, they may be reduced to one which involves only one of the unknown quantities, by any of the following methods: 1st Method. In either equation find the value of one of the unknown quantities in terms of the other and known quantities, and for it substitute this value in the other equation, which will then only contain one unknown quantity, whose VOL. I. ca-de c-b 3d Method. If either of the unknown quantities have the same co-efficient in both equations, it may be exterminated by subtracting, or adding, the equations, according as the sign of the unknown quantity, in the two cases, is the same or different. Let (x+y=152 To find x and y. By subtraction, 2 y=8, and y=4 If the co-efficients of the unknown quantity to be exterminated be different, multiply the terms of the first equation by the co-efficient of the unknown quantity in the second, and the terms of the second equation by the co-efficient of the same unknown quantity in the first; then add, or subtract, the resulting equations, as in the former case. P From certain quantities which are known, to investigate others which have a given relation to them, is the business of Algebra. When a question is proposed to be resolved, we must first consider fully its Then substimeaning and conditions. tuting for such unknown quantities as appear most convenient, we must proceed as if they were already determined, and we wished to try whether they would answer all the proposed conditions or not, till as many independent equations arise as we have assumed unknown quantities, which will always be the case, if the question be properly limited; and by the solution of these equations, the quantities sought will be determined. Prob. 1. To divide a line of 15 inches into two such parts, that one may be threefourths of the other. Let 4x one part, then 3x the other, 7x= 15, by the question, |