and 2 3x Ans. x=-3, or 1(1±√10). Ans. x=1, or (−1±√—7). Ans. x=2, or (1±√5). Ans. x=3, or (−3±√−3). + 2 = ±3 / + !J• x+7x3=22. = X +71-22: Ans. x=8, or 29±7 √—10. 2 x+7x+_22=(x—8)+7(x-2).—2 is a divisor. 8. x2+13ï3—39x=81.= x2 - 81 + 1 = (x2 = 9) x = 0 (x2 + y) { x = q) + 1 } (z 2 9) x = 0 Ans. x=3, or (-13+/-155. An artifice that is frequently employed, consists in adding to each side of the equation, such a number or quantity as will render both sides perfect squares. 9. Given, x=12+8√, to find x. x-5 Clearing of fractions, x2-5x=12+8√x. Add x+4 to each side, and extract the square root. x-2=±(4+√x). From which we easily find x=9, 4, or (−3±√7). 11. +40x+x x 49x2, 48 21=+12. + + -49=9+ 6 Add to each side. 8 x2 the Ans. x=2,-, or (-3/93) Ans. x=±2, -8, or -1. 2 Transpose 34x, and add ( 17 ) 2 13. x1 ( 1+1 ) 2—(3x2+x)=70. 3x to each side. Ans. x=2, -14, or (-2±√266). We shall now present a few solutions giving examples of other artifices. 1+x^=a(1+x)=a(1+4x+6x2+4x3+x^), (1—a)(1+x1)=4a(x+x3)+6ax2, dividing by x2, (1—a) ( x2+12 ) =1a (x+1)+6a, Complete the square, and find the value of x+, which is 2a±√2(1+a), call this 2p, and we then find x=p±√p2-1. 1-a 17. x*+=y1a, and y2+=x2, to find x and y. Whence, y=(-1±√8a+1), and x=3(4a+1=√8a+1). When two unknown quantities are found in an equation, in the form of x+y and xy, it is generally expedient to put their sum x+y=s, and their product xy=p. 55, 18. Given (x+y)(x+1)=18xy (x2+y2)(x2y2+1)=208x2y2 (2), to find x and (1), Let x+y=s, and xy=p, then s(p+1)=18p (1), y. 243 and (s2-2p)(p2+1)=208p2 (2). From the square of (1) take (2), and after dividing by 2p, we S ..s=4√√6p-18, or s2-4s/6p-18p, from S which, s=36p, or 6p. But, p+1=s, =3√6p, or √6p, Whence, p=26±√675, or 2±√3, and s=±3√{6(26±√675)}, or ±3√{6(2±√3)}. Having x+y, and xy, the values of x and y are easily found (Art. 246); two of the values are x=7±4√3, y=2F√3. { A similar substitution may be used in solving the following example : 19. 2(x+y)3+1=(x2+y2)(xy+x3+y3) (1), Ans. x=1{1±√1—8a±√2±2(1—8a)*+8a}. 22. x+y+xy(x+y)+x2y2=85, xy+(x+y)2+xy(x+y)=97. Ans. x=6, or 1. y=1, or 6. ART. 254. Two quantities of the same kind, may be compared in two ways: 1st. By considering how much the one exceeds the other. 2nd. By considering how many times the one is contained in the other. The first method is termed comparison by Difference; the second, comparison by Quotient. The first is sometimes called Arithmetical ratio, the second, Geometrical ratio. If we compare 2 and 6, we find that 2 is four less than 6, or hat 2 is contained in 6 three times. 4 Also, the arithmetical ratio of a to b is b-a, the geometrical b ratio of a to b is The term Ratio, unless it is otherwise stated, always signifies geometrical ratio. ART. 255. Ratio is the quotient which arises from dividing one quantity by another of the same kind. Thus, the ratio of 2 to 6 is 3, and the ratio of a to ma is m. ART. 256. When two numbers, as 2 and 6, are compared, the first is called the antecedent, and the second the consequent. An antecedent and consequent, when spoken of as one, are called a couplet. When spoken of as two, they are called the terms of the ratio. Thus, 2 and 6 together form a couplet, of which 2 is the first term, and 6 the second term. ART. 257. Ratio is expressed in two ways: 1st. In the form of a fraction, of which the antecedent is the denominator, and the consequent the numerator. Thus, the ratio of 2 to 6 is expressed by ; the ratio of a to b, by b 2nd. By placing two points vertically between the terms of the ratio. Thus, the ratio of 2 to 6, is written 2:6; the ratio of a to b, a:b, &c. ART. 258. The ratio of two quantities, may be either a whole number, a common fraction, or an interminate decimal. Thus, the ratio of 2 to 6 is §, or 3. The ratio of 10 to 4 is, or . The ratio of 2 to 5 is √5, or 2.236+, or 1.118+. 2 2 We see, from this, that the ratio of two quantities cannot always be expressed exactly, except by symbols; but, by taking a sufficient number of decimal places, it may be found to any required degree of exactness. ART. 259. Since the ratio of two numbers is expressed by a fraction, of which the antecedent is the denominator, and the consequent the numerator, it follows, that whatever is true with regard to a fraction, is true with regard to the terms of a ratio. Hence, 1st. To multiply the consequent, or divide the antecedent of a ratio by any number, multiplies the ratio by that number. (Arith., Part 3rd, Arts. 142, 145.) |