Complete the square, æo+3x+2=54+2=225 Some bees had alighted upon a tree; at one flight the square root of half of them went away; at another ths of them; two bees then remained. How many then alighted on the tree? (a› Let 2x=the N° of bees; Qu. 7. To divide the number 33 into two such parts, that their product shall be 162. ANSWER, 27 and 6. Qu. 8. (*) This question is taken from Mr. STRACHEY'S Translation of the Bija Ganita; and the several steps of the operation will, upon comparison, be found to accord with the Hindoo method of solution, as it stands in that Translation, page 62. Qu. 8. What two numbers are those whose sum is 29, product 100? and Qu. 9. The difference of two numbers is 5, and 4th part of their product is 26. What are the numbers? ANSW. 13 and 8. Qu. 10. The difference of two numbers is 6; and if 47 be added to twice the square of the lesser, it will be equal to the square of the greater. What are the numbers? ANSW. 17 and 11. Qu. 11. There are two numbers whose sum is 30; and d of their product plus 18 is equal to the square of the lesser number. What are the numbers? ANSW. 21 and 9. Qu. 12. There are two numbers whose product is 120. If 2 be added to the lesser, and 3 subtracted from the greater, the product of the sum and remainder will also be 120. What are the numbers? ANSW. 15 and 8. Qu. 13. A and B distribute 1200l. each among a certain number of persons: A relieves 40 persons more than B, and B gives 5. apiece to each person more than A. How many persons were relieved by A and B respectively? ANSW. 120 by A, 80 by B. Qu. 14. A person bought a certain number of sheep for 120l. If there had been 8 more, each sheep would have cost him 10 shillings less. How many sheep were there? ANSW. 40. Qu. 15. A person bought a certain number of sheep for 571. Having lost 8 of them, and sold the remainder at 8 shillings a head profit, he is no loser by the bargain. How many sheep did he buy? ANSW. 38. Qu. 16. Qu. 16. A and B set off at the same time to a place at the distance of 300 miles. A travels at the rate of one mile an hour faster than B, and arrives at his journey's end 10 hours before him. At what rate did each person travel per hour? ANSW. A travelled 6 miles per hour. B.... . 5... XXIII. On Quadratic Equations having Impossible Roots. 85. In the solution of the adfected quadratic equation, x2+px=q (Art. 81.) the two values of x were shewn to be equal to ±√p2+4q=p 2 If q be a negative quantity, and p1 less than 49, then the quantity p2-4 q is negative, and con sequently the quantity ±√p-49 comes under the description of the radical quantities mentioned in Art. 58. In this case, the two roots, or values of x, are said to be impossible. Ex. 1. Let x2+8x+31=0, or x2+8x=-31. Complete the square, (RULE I.) then x+8x+16=-31+16=-15, and x+4=-15, or x=-4-15. Ex. 2. Let x-12x+50=0, or x2 - 12x=—50. Complete the square, (RULE I.) x-12x+36=-50+36=-14, and x-6=√14; ..x=6±√−14. Ex. 3. To divide the number 16 into two such parts, that their product shall be equal to 70. Let Let x=one part, then 16-x=the other part. Hence x(16-x) or 16x-x=70. Complete the square, x2-16x+64=−70+64=-6, ..x-8=±√−6, or x=8±√−6.(«) Ex. 6. To divide the number 20 into two such parts, that On the Solution of Quadratic Equations of the form x2+px"=q. 86. Let y=x", then (by CASE III. Art. 67.) y2=x1"; and substituting these values for " and " in the equation. x2+px"=q, it is transformed into y2+py=q, where the value of y may be determined by the foregoing Rules. Having the value of y, the value of x may be found; for x"=y, ..x="/y. We are thus enabled to solve equations in which the unknown quantity is found only in two terms, (and where the index of the highest power is double the index of the lowest), like common quadratics. Ex. 1. (a) It is very well known that the greatest product which can arise from the multiplication of the two parts into which any given number may be divided, is when these parts are equal: the greatest product therefore, which could arise from the division of the number 16 into two parts, is when each of them is 8; hence, in requiring 66 to divide the number 16 into two such parts that their product should be 70," the solution of the question is impossible. These equations are often solved by the common Rules, without the formality of substitution; thus, Complete the square, (RULE I.) x−2x3+1=48+1=49. Extract the root 3-1/49=7, By RULE II. 16y'-56y+49=792+49=841, and 4y-7/841=29, or 4y=29+7=36, and y=9; ..x=y=81. Ex. 4. To resolve the number a into two such factors, that the sum of their nth powers shall be equal to b. (a) By factors are here meant the two numbers which being multiplied together shall generate the given number; if therefore x=one |