320 miles apart. A travelled 8 miles a day more than B, and the number of days before they met, was equal to half the number of miles that B travelled in 1 day. How many miles did each travel per day? Ans. A, 24 miles; B, 16 miles. 15. A passenger and freight train set out at the same time, the former from New York, and the latter from Albany, distant from each other 144 miles. The pas senger train arrived in Albany two hours after they met, and the freight train arrived in New York 8 hours after they met. At what rate did each run? Ans. 24 and 12 miles. 16. A regiment was ordered to furnish 216 men for duty, by detailing the same number of men from each company. But three companies having been detached, the remaining ones had to furnish each 12 men more to make up the required number. How many companies were there in the regiment? Ans. 9. 17. Two partners, A and B, gained 360 dollars. A's money was in trade 12 months, and he received, for principal and profit, 520 dollars. B's money was 600 dollars, and was in trade 16 months. had A? How much capital 18. A and B travel, at the same rate, towards New York. At the 50th mile-stone from New York, A overtakes a flock of geese, travelling 1 miles an hour, and 2 hours afterwards meets a stage-coach, travelling 21 miles per hour. B overtakes the geese at the 45th mile-stone, and meets the coach 40 minutes before reaching the 31st mile-stone. What is the distance between A and B? SOLUTION. Let x denote the rate of A and B's travel, and sup pose the circumstances of the problem to commence when A is at the 50th mile-stone. When B overtakes. the geese, he will have travelled 3 miles, 5 of which coincide with 5 of A's miles, Hence, the distance between A and B, is 10x 5 miles. 3 A meets the coach at the (50 — 2x) mile-stone, and B meets it at the (31 + 2) mile-stone; the coach travels, meantime, (19) miles. Hence, dividing by 32x 228 the rate of the coach, we have, the number 3 A and B. 27 8 of hours between the meetings. Adding to 19, the 3 distance that A travels in that time, we have, 8x 57 32x2 228x + miles, for the distance between 27 Hence, from the conditions of the problem, = 10x15 x2 + 2px Չ 25, the number of miles GENERAL PROPERTIES OF EQUATIONS OF THE SECOND DEGREE. 146. It has been shown that every equation of the second degree can be reduced to the form of, (1.) Transposing q to the first member, then adding p2 to and subtracting it from, the first member, which may be done without changing its value, we have, (x2 + 2px + p2) — (q + p2) = 0. (2.) Factoring the first member, on the principle that the dif ference of the squares of two quantities is equal to the sum of the quantities multiplied by their difference, we have, (x + p + √ q + p2) × (x+p−√q + p2) = 0 (3.) Equation (3) may be satisfied in two ways, and only in two ways, viz.: we may make the second factor of the first member equal to 0, and we may make the first factor equal to 0. Placing the second factor equal to 0, we have, x + p¬√q + p2 = 0. .. x = − p + √ q+p2 (4.) Placing the first factor equal to 0, we have, x+p+ v? + p2 = 0. ... xc * − p − √ Į + p2 (5.) These two suppositions give the two roots already found, and those only. Hence, we deduce this principle: 1. Every equation of the second degree has two roots, and only two. If we examine Equation (3), we see that its first member is composed of two factors of the first degree, with respect to x, the first terms of each factor being the unknown quantity, and the second terms being the two roots, each taken with the contrary sign. Hence, we deduce the following principle: 2. If all the terms be transposed to the first member, that member may be resolved into two factors of the first degree, with respect to the unknown quantity; the first term of each factor being the unknown quantity, and the second term being the two roots, each taken with its sign changed. The reverse of this principle, enables us to form an equation, when its roots are given, for which operation we may write the following EXAMPLES. RULE. Subtract each root from the unknown quantity, and multiply the results together, placing the product equal to 0. 1. Required the equation whose roots are 2 and 3. By the rule, (x − 2) (x+3) 0, whence, x2 + x 60; or, x2 + x = 6. = 2. Required the equation whose roots are 5 and 3. ― 10 3 and b. =-ab Ans. x2 Ans. x2 3 ― 3. Required the equation whose roots are a 7 58 21 13 Ans. x2 x = 3 6. Required the equation whose roots are 4 x = 6. Required the equation whose roots are -7 and —3 Ans. x2+10x 21 5 and 15 Let us resume the consideration of Equation (1), and its roots: Adding the two roots together, we have, (− p + √q + p2) + ( − p − √ 4 + p2)= −2p. Multiplying the two roots together, we have, (− p +√g + p2) × (− p −√q + p2) = −q. From these results, we deduce the following additional principles: 3. The algebraic sum of the two roots is equal to the coefficient of the second term, with its sign changed. 4. The product of the two rools is equal to the second member, with its sign changed. It follows, from the last principle, that where the second member is negative, the product of the roots will be positive; hence, the roots will have like signs. The sum of the roots, in the present case, is denoted by - 2p. Let d denote their difference: then (page 108), This product will be the greatest possible, when d = 0; herce, the greatest possible product of the two roots is when they are equal: for, in every other case, it will be less than p'. Hence, we deduce the following principle: 5. When the second member is negative, it can never be numerically greater than the square of half the coeffi cient of the second term. |