me as much as 8 yards of silk, and the number of shillings that one yard of cloth cost me, was to fivesixths of the number of yards of cloth as 4 to 5. How many yards of each did I buy? Ans. 12 yards of cloth, and 15 of silk. 9. Find two numbers whose sum is to the difference of their squares as 1 to 3, and four times their product is 72. Ans. 6 and 3. is is 10. Find two numbers the sum of whose reciprocals and the sum of the reciprocals of their squares 2 36 Ans. 3 and 6: 11. The difference of the reciprocals of two numbers is and the difference of the numbers multiplied by 2 15' twice their product is equal to 60. Required the numbers. Ans. 5 and 3. 12. The difference of the cubes of two numbers is 98, and the difference of the numbers multiplied by their product is 30. Required the numbers. Ans. 5 and 3. 13. There is a number consisting of two digits, which being multiplied by the right-hand digit, the product is 175, but if the number be divided by the right-hand digit, the quotient will be 7. Required the number. Ans. 35. 15. Five boys, A, B, C, D, and E, played at marbles; A won a certain number, B twice as many, C a number equal to twice the product of A's and B's, D as many as C, twice as many as B and one over, and E won a number equal to the square root of D's: the whole number won was 401 + 5 times A's added to D's. Required the number won by each. Ans. A won 10, B 20, C 400, D 441, and E 21. 16. Required four numbers in the proportion of the sum of whose squares is 585. 2 3' and Ans. 18, 12, 9, and 6. 17. Two merchants, A and B, entered into partnership; A's stock was in proportion to B's as 4 to 5; at the expiration of one year, A found his share of the gain to be in proportion to one-eight of his stock as 16 to 25: they continued their trade during as many years as were equal to one-fifth of the number of pounds which B contributed more than A, and at the end of that time found their whole gain to be £1440. What did each contribute to the stock, and how many years did they continue in business? Ans. A's stock £400, B's stock £500, and they continued their business 20 years. 18. The sum of two numbers is 20, and the sum of their squares 272. Required the numbers. Ans. 4 and 16. 19. Required the side of a tetraedron, the solidity of which is equal to that of a cube whose side is 2. Ans. 4.0793. 20. The difference of the sides of two cubical blocks of marble is 2 feet, and the difference of their contents 218 cubical feet. Required a side of each. Ans. 7 and 5. 111 ADFECTED QUADRATIC EQUATIONS. An Adfected Quadratic Equation is that which contains both the unknown quantity and its square. Thus, x2+6x+6=22, and 4x2+5x=26 are Adfected Quadratic Equations. But the Author of the present Treatise has considered all equations as adfected quadratics, in which there are two terms involving the unknown quantity, or any function of it, where the index of one is double that of the other. Thus, 2x1—4x2-16, x3—x3—a, x—x3—b, x‡+x3=c, x+4+√/x+4=12, are considered Adfected Quadratic Equations. Also, in equations where more than one unknown quantity is found forming two terms, one term being square of the other. Thus, (x+y)2+(x+y)=a, or x+y+√x+y=b is considered an Adfected Quadratic Equation. In the first of these equations, the first term, (x+y), is square of the second term, (x+y). Also, in the second equation, the first term, x+y is square of the second term, √x+y. RULE. 1. Transpose the unknown quantities to one side of the equation, and the known quantities to the other, observing to begin with the highest power of the unknown quantity. Then, if the first term contain a coefficient, divide the whole equation by that coefficient. 2. Complete the square, by adding the square of half the coefficient of the second term to each side of the equation. 3. Extract the square root of each side of the equation, and the unknown quantity will be easily determined as follows: 1. In the first example, viz. x2+6x+6=22. 2. Transpose the 6, and the equation becomes x2+6x=16. 3. Here, the unknown quantity in the first term is square of the unknown quantity in the second term. 4. Then 6 being the coefficient of the second term, half of which is 3, this being squared and added to each side of the equation, it becomes x2+6x+9=16+9=25. 5. Extracting the square root of each side of this equation, and it becomes x+3=±5; 6. by transposition, x=5-3=2, or -8. In the second example, 4x2+5x=26. 1. Here, the first term has a coefficient, 4. 2. Divide the whole equation by 4, and it becomes 5 term, half of which is, this being squared and added to each side of the equation, it be ADFECTED QUADRATIC EQUATIONS. 113 4. Extracting the square root of each side of this In the third example, 2x-4x2=16. 1. Here, the index of the unknown quantity in the first term is double that of the second, and the first term has a coefficient. 2. Divide by 2, the coefficient of the first term, x2-2x2-8. 3. Complete the square, -2x2+1=8+1=9. 4. Extract the square root, x2-1=±3. 5. By transposition, a2+3+1=4, or — 2. 6. Extracting the square root, x=±2, or ±√-2. In the example, x+4+x+4=12. 1. Here, x+4 is considered the first term, and x+4 the second term, .. the first term is square of the second. 2. One being the coefficient of the second term, half of which is 3. Completing the square, x+4+√x+4+ — 1 49 12+ == 4. Extracting the square root, √x+4+ 5. By transposition, √x+4= 6. Squaring each side, x+4=9. 7. By transposition, x=5. 1 7 3. |