13th. +3x+10 14th. x2-8x=14?.............. Ans. x=9·4772+ ; x=−1·4772+. 15th. 3x2+x=7?............... Ans. x=1·3699+; x=−1·7032+. 16th. 6x-30=3x2?..... ....Ans. x=1+3=1. ac 17th. cx+a+b=(a+b) x2?...Ans. x= 2(a+b) 'a-b Ans. x=± √a+b; y=±√ 2 19th. a±√2b- a2 a√2b-a. ; y= 20th. 21st. Ans. (x=±√/b+2a±√/b−2a or x=±√ 1b+b2-4a2 2 y=±√b+2a+√/b—2a or y=± √ b√/b2-4a2. ................................Ans. x=±3; x=±2. 22d. x1-13x2+36=0?......... 23d. (x2-2)2=4(x2+12) ?.....................................................Ans. x=±2; x=±}. 24th. (x2—1)(x2—2)+(x2—3)(x2—4)=x1+5? Ans. x+1; x=±3. 25th. It is required to find a number such that, if its half is multiplied by its third part, the product shall be 864? Ans. 72. 26th. Find two numbers whose product is 750, and the quotient obtained by dividing the greater number by the ...Ans. 50 and 15. less 3?........ 27th. The product of two numbers a, their quotient =b; required the expressions of the numbers in terms of a, b?............ ........Ans. √ab and √%. 28th. The sum of the squares of two numbers is 13001, and the difference of their squares is 1449; required the num..Ans. 85 and 76. bers ?.......... 29th. The sum of the squares of two numbers being expressed by a, and the difference of their squares by b, it is required to express the two numbers in terms of a, b? Ans. 30th. If two numbers are to each other as 3 to 4, and the sum of their squares is 324900, what are the numbers? Ans. 342 and 456. 31st. The sum of the squares of two numbers is b, and the first number is to the second as m is to n; what are the mb √ m2 + n23 √ m2 + n2° 32d. The difference of two numbers is 8 and their product 240; required the numbers?............... Ans. 12 and 20. 33d. The difference of two numbers is a, and their product b; required the numbers? 34th. Find a number whose square exceeds the number itself by 306?.... ...................................Ans. 18. 35th. A charitable person distributes 36 £. in equal shares amongst the poor of a village; six of those whom he intended to relieve needing no assistance, each of the remaining paupers receives 10 sh. 8 d. more than he otherwise would have had; how many paupers were there at first ?......................... .......Ans. 54. 36th. A horse-dealer pays a certain sum for a horse, which he afterwards sells for 144 £., and gains exactly as much per cent. as the horse cost him; how much did the horse cost?........... .........Ans. 80 £. 37th. The sum of two numbers being 41, and the sum of their squares 901; what are the numbers ?....Ans. 15 and 26. 38th. Divide the number 16 into two parts, such that the squares of the two parts augmented by their product shall be 208?......... ....................Ans. The parts are 4 and 12. 39th. A grazier bought a certain number of oxen for 240 £.; after losing three he sold the remainder for 8 £. a head more than they cost him, and thus gained 59 £. by the bargain; what number did he buy?.. ............ Ans. 16. 40th. It is required to find two numbers such, that if they are multiplied respectively by the given numbers a, b, the sum of the products shall be a given number, p; and if their squares are multiplied respectively by the same numbers a, b, the sum of the new products shall be equal to another given number, q. Ans. x= ap±√ab[(a+b)q—p2]y_bp+✔ab[(a+b)q−p2] a(a+b) b(a+b) SECTION VII OF THE FORMATION OF POWERS AND EXTRACTION OF ROOTS. 131. Let abc.. . . be a product, and n any positive whole number. (abc . . . .)”=abc Xabc....x abc. ... . &c. to n repetitions of the quantity abe =aaa... to n repetitions of the factor ax bbb.... to n repetitions of the factor bxccc to n repetitions of the factor cx =a"b"c Whence a product is raised to any power by raising each of its factors to that power. Let it be required to raise aTM to the power n, m and n being whole and positive numbers. (a")"=a”×a×a":.. to n repetitions of the quantity aTM=am+m+m... Consequently a quantity which is affected with an exponent is raised to a power by multiplying its exponent by the degree of that power. If it is required to raise any monomial to a power, regarding the monomial as a product, all its factors must be raised to the required power. This is accomplished by raising the coefficient to the required power, and multiplying all the exponents by the degree of that power. When a quantity is positive, all its powers are positive; when negative it must have + before the second power, before the third, + before the fourth, &c. (Art. 4. c). Whence, in general, whatever be the sign of a quantity, the even powers of that quantity have the sign +, and the odd powers have the same sign as the quantity. Hence (+2a3b5c)+=+16a12b20c4. 132. By reversing these rules the following rules for the extraction of roots are obtained: 1st. The root of a product is obtained by extracting the required root of each factor of that product. 2d. The root of a quantity, which has an exponent, is obtained by dividing that exponent by the exponent or index of the root. 3d. The root of any monomial is obtained by extracting the root of the coefficient, and dividing the exponent of each letter by the index of the root. 4th. When the index of the root extracted is even, the root ought to have the ambiguous sign, †; and when the index of the root is odd, the root has the same sign as the power. 133. When the coefficient of a monomial is not an exact power of the order marked by the index of the root which is to be extracted, or when the exponents of the different factors are not divisible by that index, the root of the monomial is first indicated by means of the sign; then the radical is simplified, as has been already done in the case of extraction of the square root. As an illustration let it be required to extract the 5th root of 96ab5c12. 96-25×3; a2=a5 × a2; cl2=c10 X c2. .../96a2bc12=25a5b5c10 × 3ac2abc2 √3a2c2. Whence to simplify any radical it is necessary to decompose the quantity under the radical sign into two factors, of which the one involves exact powers of the same order as the root which is to be extracted, and the other involves no exact powers of that order; then to extract the root of the first factor, and to indicate the root of the second. 134. No real quantity, whether positive or negative, when raised to an even power, can give a negative result. Consequently every expression composed of a radical of an even degree placed over a negative quantity represents an imaginary quantity. Supposing a, b, positive magnitudes, the expressions -2abb, are imaginary. -a, -a3, But expressions such as these may, like imaginary radicals of the second degree, be transformed into products in which the only imaginary quantity is a root of -1. For example, a=Vax-1=Väx√=1. 135. In the same manner as division has led to the use of negative exponents, the extraction of roots leads to the use of fractional exponents. By the preceding rule the 5th root of al is obtained by dividing the exponent of a by 5, and the root required is a2. But the division may be indicated only, and the expression written either thus, Val, or thus, a. If the exponent of a is not divisible by 5, the root cannot be extracted, but nothing prevents the indication of the division of the exponent, which in such a case obtains for denominator the index of the root. This convention has been adopted by algebraists, and in consequence the two expressions /a" and a are considered equivalent. THEORY OF ARRANGEMENTS, PERMUTATIONS, AND COMBINATIONS. 136. To form any power whatever of a binomial, it is incidentally found to be necessary to investigate the theory of arrangements, permutations, and combinations. Many letters, a, b, c, d, e...being given, suppose them taken two by two in all manners possible, in such sort that each assemblage of two letters differs from the others either by the letters which compose it, or by the order in which they are written; there are thus formed what are named the arrangements, two by two, of the given letters. When the letters are, in like manner, taken three by three, four by four, &c., the arrangements three by three, four by four, &c. are obtained. It is easy to form these arrangements and to determine their number. In forming them it is convenient to follow an order by which no one of the terms will be repeated. Thus in arranging the terms two by two it is sufficient to annex in turn to each letter all the other letters. In this manner are obtained arrangements such as ab, ac, ad, ae, af, ca, cb, cd, ce, cf, and it is manifest, if the number of letters a, b, c, d..... is m, and the number of arrangements two by two is A, that each horizontal row contains one arrangement fewer than there are letters (or m-1), and that there are as many horizontal rows as there are letters (m); and therefore that the whole number of arrangements, or A„=m(m—1). The arrangements three by three of the letters a, b, c, d, e, . . . . can be formed from the arrangements of the letters, taken two by two, by annexing to each of the m(m-1) arrangements in turn each of the m-2 letters which do not enter into the arrangement. Thus from ab are formed abc, abd, abe, abf,.. from ac .... acb, acd, ace, acf, If these arrangements are represented at length in a tabular form, it is evident that there are A, horizontal rows, and that the number of arrangements in each row is m-2. Therefore, making A, the expression of the whole number of arrangements of m letters, taken three by three, A, A2X (m-2)=m(m-1)(m-2). The arrangements four by four of the letters a, b, c, d, e.... are formed from the arrangements of the same letters taken three by three, by annexing to each of the m(m-1)(m-2) arrangements in turn each of the m-3 letters which do not enter into the arrangement. Denoting by A, the whole number of arrangements of m letters, taken four by four, A=A,× (m-3)=m(m−1)(m−2)(m—3). For every additional letter introduced into the arrangements, it is evident that another factor must be introduced into the formula, and that this factor must be less by 1 than the factor which precedes it. Whence the general expression of the number of arrangements of m letters, taken n by n, must involve n factors, taken consecutively in the series m, m-1, m-2, &c., the last factor being m— —(n−1) or m—n+1. Consequently, denoting by A, the number of arrangements of m letters, taken n by n, A=m(m-1)(m-2) (m-3)....... (m-n+1) 1 If n is made successively 2, 3, 4,...the last factor m-n+1 becomes m-1 in the first instance, m-2 in the second, m-3 in the third;. and the values of A2, A3, A4, · are recovered. .... ...... ...... 137. If any letters, abcde.. are written consecutively, and the order of these letters is changed in every possible manner, their permutations are obtained. For example, the permutations of the three letters abc are, abc, acb, bac, bca, cab, cba. The permutations of n letters are consequently the arrangements of these n letters taken n by n. Whence, if m is replaced by n, and the number of permutations of n letters is denoted by P,, formula 1 gives P=n(n-1)(n-2)(n−3). ....... or P=n(n-1)(n−2)(n−3).. or, by reversing the order of the factors, P=1×2×3×......xn ... .(n−n+1), 2 Formula 2 can be obtained as follows, without deduction from formula 1 : With two letters only two permutations, ab, ba, can be formed. The permutations of three letters are obtained by writing successively after each of the three letters the permutations of the two others; this gives 3×2 or 2×3 as the number of permutations which can be formed with three letters. Similarly for 4 letters the number of permutations is 2×3×4, and for n letters it is 1×2×3×.....xn. 138. Of the arrangements of m letters, taken n by n, some differ from others only in respect that the same letters are taken in different orders, as abc, acb, bac, bca.....; while others differ from each other in respect that, although the number of letters is the same, one or more of these letters are different, as abc, abd, acd, bcd.... The latter arrangements are distinguished by the name of Combinations or Products. It is to be observed that the six arrangements, abc, acb, bac, bca, cab, cba, form only one combination or product (Part I. Art. 73). Among the arrangements of m letters n by n, it is evident that each combination of n letters must be repeated as many times as there are permutations of the n letters of that combination. Therefore, if the total number of arrangements which can be formed of m letters, taken n by n, is divided by the number of permutations which can be formed with n letters, the number |