more than the other. How many square rods do both gardens contain ? Ans. 208 square rods. 34. A has two square gardens, and the side of the one exceeds that of the other by 4 rods, and the contents of both are 208 square rods. How many square rods does the larger garden contain more than the smaller? · Ans. 80 square rods. 35. I have two blocks of marble which are exact cubes, and whose united lengths are 20 inches, and they contain 2240 cubic inches. Required the surface of each. Ans. Larger, 864 inches; smaller, 384 inches. 36. A merchant sold a bale of cloth for $75, and gained as much per cent. as the cloth cost him. What was the price of the cloth ? Ans. $50. 37. There are two numbers whose difference is 12, and whose sum multiplied by the greater is 560. What are those numbers? Ans. 20 and 8. 38. The plate of a looking-glass is 36 inches by 12 inches. It is to be framed with a frame all parts of which are of equal width, whose area is 448 square inches. What is the width of the frame? Ans. 4 inches. 39. Divide 100 into two such parts that the sum of their square roots shall be 14. Ans. 64 and 36. 40. A square court-yard has a rectangular gravel-walk around it. The side of the court wants one yard of being six times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the number of yards in the perimeter of the court by 340. What is the area of the court and width of the walk? Ans. Area of the court, 529 square yards; width of the walk, 4 yards. 41. A merchant bought 54 bushels of wheat, and a certain quantity of barley. For the former he gave half as many shillings per bushel as there were bushels of barley, and for the latter 4 shillings per bushel less. He sold the mixture at 10 shillings per bushel, and lost £28 16s. by his bargain. What was the price of the barley? Ans. 36 bushels of barley, at 14 shillings per bushel. 42. I have 165 square feet of plank, 3 inches in thickness, with which I intend to make a cubical box. Required its contents in cubic feet. Ans. 125 cubic feet. 43. I have a small globule of glass, one inch in diameter. How large a sphere may be made of it, if the glass is to be only of an inch in thickness, taking it for granted that all spheres are to each other as the cubes of their diameters? Ans. Inside diameter, 1.775+ inches; whole diameter, ⚫ 1.875 inches. 44. John Smith has two cubical boxes, whose united lengths in the clear are 20 inches, and their solid contents are 2240 cubic inches. What is the difference of their contents? Ans. 1216 cubic inches. 45. I have two house-lots, which contain 6100 square feet, and the larger contains 1100 square feet more than the less. Required their dimensions. Ans. 50 and 60 feet square. 46. Two men, A and B, bought a farm of 200 acres, for which they paid $200 each. On dividing the land, A says to B, If you will let me have my part in the situation which I shall choose, you shall have so much more land than I that mine shall cost 75 cents per acre more than yours. B accepted the proposal. How much land did each have, and what was the price of each per acre? Ans. A had 81.866 acres, at $2.443+; B had 118.133+ acres, at $1.693+. 47. A and B engaged to reap a field for 90 shillings. A could reap it in 9 days, and they promised to complete it in 5 days. They found, however, that they were obliged to call in C, an inferior workman, to assist them the last two days, in consequence of which B received 3s. 9d. less than he otherwise would have done. In what time could B and C each reap the field? Ans. B could reap the field in 15 days, and C in 18 days. SECTION XVIII. CUBIC AND HIGHER EQUATIONS. ART. 226. A Cubic Equation is one in which the highest power of the unknown quantity is the third power. 227. A Biquadratic is an equation in which the highest power of the unknown quantity is the fourth power. 228. An equation of the fifth degree 'is one in which the highest power of the unknown quantity is the fifth power. As, x-ax1+bx3-cx2+dx=e. And so on, for all other higher powers. There are many particular and very prolix rules given for the solution of the above-mentioned equations; but they all may be readily solved by the following easy RULE. 1. Find, by trial, two quantities as near the true root as convenient, and substitute them separately, in the given equation, instead of the unknown quantity, and find how much the terms collected together, according to their signs + or −, differ from the known members of the equation, noting whether these errors are in excess or deficiency. 2. Multiply the difference of the two quantities found, or taken by trial, by either of the errors, and divide the product by the difference of the errors when they are alike, but by their sum when they are unlike. Or, we may say, as the difference or sum of the errors is to the difference of the two assumed quantities, so is either error to the correction of its supposed quantity. 3. Add the quotient last found to the quantity belonging to that error when its supposed quantity is too little, but subtract it when too great, and the result will give the true root nearly. 4. Take this root, and the nearer of the two former, or any other that may be found nearer; and, by proceeding in like manner as above, a root will be obtained nearer than before. Proceeding in the same manner, we may obtain the answer to any degree of exactness required. NOTE 1. -It is best always to employ two assumed quantities, that shall differ from each other only by unity in the last figure on the right, because then the difference, or multiplier, is only 1. It is also best to use always the less error in the above operation. NOTE 2.- It will be convenient, also, to begin with a single figure at first, trying several single figures, till there be found the two nearest the truth, the one too little, and the other too great; and, in working with them, find only one more figure. Then substitute this corrected result in the equation for the unknown letter; and, if the result prove too little, substitute also the number next greater for the second supposition; but, if the former prove too great, then take the next less number for the second supposition; and, working with the second pair of errors, continue the quotient only so far as to have the corrected number to four places of figures. Then repeat the same process again with this last corrected number, and the next greater or less, as the case may require, carrying the third corrected number to eight figures, because each new operation commonly doubles the number of true figures. Proceed in this manner to any extent that may be wanted. EXAMPLES. 1. Find the root of the cubic equation x+x+x=100. We see that x lies between 4 and 5. We assume, therefore, 4 and 5 as the two values of x. |