Hence, the money of A. was 79+ crowns. 4. Some merchants appoint a factor at Archangel. Each contributes to the trade ten times as many crowns as there are partners in the concern. The profit of the factor is fixed at twice as many crowns per cent. as there are partners. Now if part of his total gain be multiplied by 2%, the product will give the number of partners. Required the number of partners, and the sum contributed by each. Assume for the number required. Then since by the question each partner contributes 10r crowns, the whole capital will be 10x2. Again by the question, for every 100 crowns the factor gains for his profit 2x crowns, so that 100: 2x:: 10x2: 1 20x3 1 3 his whole profit. Now 2 Now part of×2 gives the number of partners, 5 which was assumed to be x. We have therefore the following equation, Dividing by x, x2 = 225, and x = 15. Therefore the number of partners was 15, and each contributed 150 crowns. 5. In a right-angled triangle the hypothenuse is 13 feet, and the difference of the sides is 7 feet. Find the length of the sides. Let x and y = the sides. Then (Euclid. Prop. 47. B. 1.) x2 +y2=169. And by the question, x —y=7. •. x2 -2xy +y=49 6. Find two numbers in the ratio of 5: 3, and the differ ence of whose squares shall be 144. ANS. 15 and 9. 7. The sum of two numbers is 10, and the sum of their squares is 52. What are the numbers? 8. Divide 89 into two such parts that the sum of their square roots may be 13. ANS. 64 and 25. 9. There are two numbers, whose difference is to the difference of their square roots as 8: 1, and the square root of whose product is 15. What are the numbers? ANS. 25 and 9. 10. The difference of the cubes of two numbers is 56, and their product multiplied by their difference is 16. Required the numbers. ANS. 4 and 2. 11. A number of persons engaged in play, each bringing as many purses as there were persons; and each purse containing 6 times as many pounds as there were persons. Now the whole number of pounds being 384, how many persons were there? ANS. 4. t 12. In a division sum the quotient and remainder were both found to be 23; and the square of their sum was less than the square of the divisor by 1020. Required the divisor and dividend. 212. A CHAP. XXIV. OF ADFECTED QUADRATIC EQUATIONS. QUADRATIC Equation is said to be Mixt, or Adfected, when it contains all the three terms mentioned in the preceding chapter; that is to say, when it contains the square of the unknown quantity, as ax2; the first power of the unknown quantity, as be; and the term which is composed of known quantities only. Now since we may unite two or more terms of the same kind, and bring every term to one side of the equation, the form of an adfected quadratic will be, ax2±bx±c=0. We will now proceed to show how the value of x may be determined from an equation of this nature; and it will presently appear that there are two methods of arriving at it. 213. Now an equation of this kind may be first reduced by means of division to a form where the first term contains simply the square of the unknown quantity, that is to say, without having any co-efficient, and which will then be of the form of ს C a 0. And, assuming a to be a measure of b and c, it will be thus reduced to the form of, and by transposing q, the known quantity, to the other side of the equation, we shall have, where Ρ and q represent any whole numbers whatever, either positive or negative. To proceed therefore with the solution of the equation. We shall commence by observing, that if x2+px were a complete square, no difficulty could occur in arriving at the value of x, since it would evidently be at once determined by extracting the root of each side of the equation. 214. It is evident, however, that x2±px cannot be a complete square, since we have already seen, Chap. XVIII, that where the root consists of 2 terms, as x+n, the square will always consist of 3, that is, of the square of each term, and of twice their product, viz. x2+2nx +n2. Now since we have already got x2+px, we may therefore consider 2 as the square of the first term of the root; and in that case pr will represent twice the product of the first term a multiplied into the second; it is evident therefore that the second term will be ; the square of which, viz., being added, will render x2±px a complete square; and it will be seen immepe diately by trial that x2+px + is the exact square of x±1p. 4 215. Since therefore it has been demonstrated that p2 x2±px + is a complete square, having for its root xp, 4 if we resume our equation, x2+px==q, it is clear that we have only to add to each side of it the or, and it square of half the coefficient of the second term, or " 4 of which the first side is a complete square, and the second contains only known quantities. Now by extracting the roo of each side we shall have, by transposition, x=={p±√‡•p2 Fq; or 2 216. Such |