Here, agreeably to the above observation, let x = vy, then v2y + vy2 = 56, and vy2 + 2y2 = 60, whence, from the v+2 or 60υ2 + 60v = 56v + 112. And, by transposing 56v, and (18) = 60 60 v+21++2 4 = 18, or =X3√2 = 4√2. y= ✓ 2. The work 3 √2, and x = vy = may also be sometimes shortened, by substituting for the unknown quantities, the sum and difference of two other quantities; which method may be used when the unknown quantisies, in each equation, are similarly involved. Here, according to the above observation, let there be assumed x = z + v, and y = z - v. Then, by adding these two equations together, we shall have x + y = 2z = 12, or * = 6, also, since x = 6 + v, y = 6 - v, and by the first equation x3 + y2 = 18xy, we shall obtain, by substitution, (6 + v)3 + (6 – υ)3 = 18 (6 + υ) (6 – υ), or, by involving the two parts of the first member, and multiplying those of the second, 432 +3602 = 648 - 18v2, whence, by transposition, 5402 = 216; and by division, v2 = 216 4; or v = ±2. And therefore, by the first assumption, and the first part of the process, we have x = z + v = 6 ±2=8, or 4, and y=z- v = 6 ±2 = 4, or 8. QUESTIONS PRODUCING QUADRATIC The methods of expressing the conditions of questions of this kind, and the consequent reduction of them, till they are brought to a quadratic equation, involving only one unknown quantity and its square, are the same as those already given for simple equations. 1. To find two numbers such that their difference shall be 8, and their product 240. Let x equal the least number. Then will x + 8 = the greater, And x (x + 8) = x2 + 8x = 240, by the question, Whence x = -4 + √(16+240) = 4+256 by the common rule, before given, Therefore x = 16 - 4 = 12, the less number, and x + 8 = 12 + 8 = 20, the greater. 2. It is required to divide the number 60 into two such parts, that their product shall be 864. Let x = the greater part, Then will 60 - x = the less, And x (60-x) = 60x -x2 = 864, by the question, Or, by changing the signs on both sides of the equation, 60x = -864, Whence x = 30 ± √ (900 - 864) = 30 ± √ 36 = 30 ± 6, by the rule, And consequently, x = 30+ 6 = 36, or 30- 6 = 24, the two parts sought. 3. It is required to find two numbers such, that their sum shall be 10 (a), and the sum of their squares 58 (b). Let x = the greater of the two numbers, And x2 + (a - x)2 = 2x2 - 2x + a2 = b, by the question, Or 2x2 - 2ax = b - a2, by transposition, b-a by division. 2 == 2 by the rule, And if 10 be put for a, and 58 for b, we shall have == 10 1 이+√(116-100) = 7, the greater number, 10 1 And 10 2 (116 - 100) = 3, the less. 4 Having sold a piece of cloth for 341., I gained as much per cent. as it cost me; what was the price of the cloth ? Let x = pounds the cloth cost, Then will 24 - x = the whole gain, Whence x= That is x2 + 100x = 2400, 50+ v (2500 + 2400) = -50+70= 20, by the rule, And consequently, 20l. price of the cloth. 5. A person bought a number of sheep for 80l., and if he had bought four more for the same money, he would have paid 11. less for each; how many did he buy? Let a represent the number of sheep, And = price of each, if x + 4 cost 80l. Then will 80 x+4 And 80x + 320 = 80x + x2 + 4x, by the same, Or, by leaving out 80x on each side, x2 + 4x = 320, Whence x = -2 +(4+320) = -2 +18, by the rule, And consequently, x = 16, the number of sheep. 6. It is required to find two numbers, such that their sum, product, and difference of their squares, shall be all equal to each other. Let x = the greater number and y = the less. Hence 1 = } by the question. = x - y, or x = y + 1, by 2d equation. x+y And (y + 1) + y = y (y + 1), by 1st equation, That is, 2y + 1 = y2 +y; y2 + y = 1. 1 1 +√5, by the rule, √5 = 1.6180 ... 1 2 Where denotes that the decimal does not end. ... 7. It is required to find four numbers in arithmetical progression, such that the product of the two extremes shall be 45, and the product of the means 77. Let x = least extreme, and y = common difference, Then x, x + y, x + 2y, and x + 3y, will be the four numbers, = Hence x (x + 3y) = x2 + 3xy = 45 And 2y2 = 77-45 = 32, by subtraction, Or, y3 = 32 = 16 by division, and y = √ 16 = 4. Therefore, x2 + 3xy = x2 + 12x = 45, by the 1st equation, And consequently, x = -6+ √(36+45) = -6 + 9 = 3, by the rule. Whence the numbers are 3, 7, 11, and 15. 8. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Let x, y, and z, be the three numbers, Then, xz = y2, by the nature of proportion, And { x + y + z = 14 +2=84 by the question, Hence, x + z = 14 - y, by the second equation, And x2 + 2zx + z2 = 196 -- 28y + y23, by squaring both sides, Or x2 + z2 + 2y2 = 196 - 28y + y2 by putting 2y2 for its equal 2xz, That is, x2 + y2 + z2 = 196 – 28y, by subtraction, Hence y = Or, 19628y = 84, by equality, 196 84 28 = 4, by transposition and division. 16 Again, xz = y2 = 16, or x = -, by the 1st equation, And x + y + z = 16 2 +4 + z = 14, by the 2d equation, Or, 16 + 4z + z2 = 14z, or 22 10z = -16, Whence z = 5±√(25-16) = 5±3=8, or 2 by the rule, Therefore, the three numbers are 2, 4, and 8. 9. It is required to find two numbers, such that their sunm shall be 13 (a), and the sum of their fourth powers 4721 (b). Let x = the difference of the two numbers sought, 1 1 2 2 Then will a + 5x, or a + x 2 = the greater number, Or (a + x)2 + (a - x) = 166, by multiplication, And x2 + 6a2x2 = 8b - a, by transposition and division, Whence x = - 3a2 + √(9a2 + 8b √8 (a + b), by the rule, a*) = 3a2+ And x = √[-3a2 + 2√2 (a2 + b)], by extracting the root, Where, by substituting, 13 for a, and 4721 for b, we shall have x = 3, The sum of which is 13, and 84 + 5 = 4721. 10. Given the sum of two numbers equal s, and their product = p, to find the sum of their squares, cubes, biquadrates, &c. Let x and y denote the two numbers; then (1.) x + y = s, (2.) xy = p. From the square of the first of these equations take twice the second, and we shall have (3.) x2 + y2 = 2p = sum of the squares. Multiply this by the 1st equation, and the product will be x3 + xy2 + x2y + y3 = s3 2sp. From which subtract the product of the first and second equations, and there will remain (4.) x3 + y3 = s - 3sp = sum of the cubes. Multiply this likewise by the 1st, and the product will be x2 + xy2 + xy + y2 = s* - 3sp; from which subtract the product of the second and third equations, and there will remain (5.) x + y = s2 - 4sp + 2p2 = sum of the biquadrates. And, again multiplying this by the 1st equation and subtracting from the result the product of the second and fourth, we shall have 2 (6.) x + y = s5 - 5sp + 5sp2 = sum of the fifth powers. And so on; the expression for the sum of any powers in general being xm + ym = sm - msm-2p+ m(m-3) 2 sm-4p |