But 5:2 :: 2x (the whole number of eggs) : *; * + - r = 6; each sort. Quest. 10. . Two persons, A and e, engage at play. Before they begin, a has 80'guineas, and B has 60. After a certain number of games won and lost between them, A rises with three times as many guineas as B. Query, how many guineas did a win of B? Let x denote the number of guineas A won. 3.2 ; QUESTIONS FOR PRACTICE. of the se-, 1. To determine two numbers such, that their difference may be 4, and the difference of their squares 64. Ans. 6 and 10. 2. To find two numbers with these conditions, viz. that half the first with a 3d part of the second may make 9, and that a 4th part of the first with a 5th part cond may make 5. Ans. 8 and 15. 3. To divide the number 20 into two such parts, that a 3d of the one part added to a fifth of the other, may make 6. Ans. 15 and 5. 4. To find three numbers such, that the sum of the 1st and 2d shall be 7, the sum of the 1st and 3d 8, and the sum of the 2d and 3d 9. Ans. 3, 4, 5. 5. A father, dying, bequeathed his fortune, which was 28001. to his son and daughter, in this manner; that for every half crown the son might have, the daughter was to have a shilling. What then were their two shares ? Ans. The son 2000l. and the daughter 8001. 6. Three persons, A, B, C, make a joint contribution, which in the whole amounts to 4001.: of which sun B con tributes tributes twice as much as A and 201. more; and c as much as A and B together. What sum did each contribute ? Ans. A 601. B 1401, and c 2001. 7. A person paid a bill of 1001. with half guineas and crowns, using in all 202 pieces; how many pieces were there of each sort ? Ans. 180 half guineas, and 22 crowns. 8. Says A to B, if you give me 10 guineas of your money, I shall then have twice as much as you will have left: but says B to A, give me 10 of your guineas, and then I shall have 3 times as many as you. How had each? Ans. A 22, B 26. 9. A person goes to a tavern with a certain quantity of money in his pocket, where he spends 2 shillings; he then borrows as much money as he had left, and going to another tavern, he there spends 2 shillings also; then borrowing again as much money as was left, he went to a third tavern, where likewise he spent 2 shillings; and thus repeating, the. * same at a fourth tavern, he then had nothing remaining. What sum had he at first? Ans. 35. 9d. 10. A man with his wife and child dine together at an inn. The landlord charged 1 shilling for the child; and for the woman he charged as much as for the child and I as much as for the man; and for the man he charged as much as for the woman and child together. How much was that for each ? Ans. The woman 200. and the man 32d. 11. A cask, which held 60 gallons, was filled with a mixture of brandy, wine, and cyder, in this manner, viz. the cyder was 6 gallons more than the brandy, and the wine was as much as the cyder and of the brandy. How much was there of each? Ans. Brandy 15, cyder 21, wine 24. 12. A general, disposing his army into a square form, finds that he has 284 men more than a perfect square ; but increasing the side by 1 man, he then wants 25 men to be a complete square. Then how many men had he under his command ? Ans. 24000. 13. What number is that, to which if 3, 5, and 8, be severally added, the three sums shall be in geometrical progression ? Ans. l. 13. The stock of three traders amounted to 8601. the shares of the first and second exceeded that of the third by A by 240 ; and the sum of the 2d and 3d exceeded the first by 260. What was the share of each ? Ans. l'he 1st 200, the 2d 300, the 3d 260. 15. What two numbers are those, which, being in the ratio of 3 to 4, their product is equal to 12 times their sum? Ans. 21 and 28. 16. A certain company at a tavern, when they came to settle their reckoning, found that had there been 4 more in company, they might have paid a shilling a-piece less than they did; but that if there had been 3 fewer in company. they must have paid a shilling a-piece more than they did, What then was the number of persons in company, what each paid, and what was the whole reckoning? Ans. 24 persons, each paid 7s. and the whole reckoning 8 guineas. 17. A jockey has two horses ; and also two saddles, the one valued at 181. the other at 31. Now when he sets the better saddle on the 1st horse, and the worse on the 2d, it makes the first horse worth double the 2d: but when he places the better saddle on the 2d horse, and the worse on the first, it makes the 2d horse worth three times the 1st. What then were the values of the two horses ? Ans. The 1st Gl., and the 2d 91. 18. What two numbers are as 2 to 3, to each of which if 6 be added, the sums will be as 4 to 5 ? Ans. 6 and 9. 19. What are those two numbers, of which the greater is to the less as their sum is to 20, and as their difference is to 10? Ans. 15 and 45. * 20. What two numbers are those, whose difference, sum, and product, are to each other, as the three numbers 2, Ans. 2 and 10. 21. To find three numbers in arithmetical progression, of which the first is to the third as 5 to 9, and the sum of all three is 63. Ans. 15, 21, 27. 22. It is required to divide the number 24 into two such parts, that the quotient of the greater part divided by the less, may be to the quotient of the less part divided by the greater, as 4 to 1. Ans. 16 and 8. 23. A gentleman being asked the age of his two sons, answered, that if to the sum of their ages 18 be added, the result will be double the age of the elder; but if 6 be taken 3, 5? first power in another terni, taken from the difference of their ages, the remainder will be equal to the age of the younger. What then were their ages? Ans. 30 and 12. 24. To find four numbers such, that the sum of the 1st, 2d, and 3d shall be 13 ; the sum of the 1st, 2d, and 4th, 15; the sum of the 1st, 3d, and 4th, 18; and lastly the sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 25. To divide 48 into 4 such parts, that the first increased by 3, the second diminished by 3, the third multiplied by 3, and the 4th divided by 3, may be all equal to each other. Ans. 6, 12, 3, 27. QUADRATIC EQUATIONS. QUADRATIC Equations are either simple or compound. A simple quadratic equation, is that which involves the square of the unknown quantity only. As ax? = b. And the solution of such quadratics' has been already given in simple equations. A compound quadratic equation, is that which contains the square of the unknown quantity in one term, and the As ax? + bx = c. All compound quadratic equations, after being properly reduced, fall under the three following forms, to which they must always be reduced by preparing them for solution. 1. x2 + ar = b b The general method of solving quadratic equations, is by what is called completing the square, which is as follows: 1. Reduce the proposed equation to a proper simple form, as usual, such as the forms above ; namely, by transposing all the terms which contain the unknown quantity to one side of the equation, and the known terms to the other; placing the square term first, and the single power second; dividing the equation by the co-efficient of the square or first term, if it has one, and changing the signs of all the terms, when that term happens to be negative, as that term must always be made positive before the solution. Then the proper solution is by completing the square as follows, viz. 2. Complete a.r = 2. Complete the unknown side to a square, in this manner, viz. Take half the co-efficient of the second term, and square it; which square add to both sides of the equation, then that side which contains the unknown quantity will be a complete square. 3. Then extract the square root on both sides of the equation*, and the value of the unknown quantity will be determined, As the square root of any quantity may be either + or -, therefore all quadratic equations admit of two solutions. Thus, the square root of tn? is either + nor en; for tnx tn and - nx n are each equal to + n°. But the square root of - no, or ✓ –no, is imaginary or impossible, as neither + n nor - n, when squared, gives - no. So, in the first form, x2 + ax=b, where x + ja is found = vb + a2, the root may be either + vo + tu?, or -7b+fa”, since either of them being multiplied by itself produces b + a. And this ambiguity is expressed by writing the uncertain or double sign I before 7b+ Lao; thus ==Vb+ a? — {a. In this form, where x = vb.+ da? - ju, the first value of x, viz. x = + vb + bar -- a, is always affirmative; for since fa? + b is greater than ja”, the greater square must necessarily have the greater root; therefore vb+ba? will always be greater than va', or its equal ja; and consequently + Vb+ fa will always be affirmative. The second value, viz. x = Nb + ju? - ja will always be negative, because it is composed of two negative terms. Therefore when x2 + ax = b, we shall have x = + vota? - ja for the affirmative value of x, and x abt ta' the negative value of x. In the second form, where x = + vb tas + fa the first value, viz. x = + b + a + La is always affirmative, since it is composed of two affirmative terms. But the second value, viz. vb + a2 + a, will always be negative ; for since b + La is greater than Ja, therefore vba' will be greater than va?, or its equal ja ; and consequently - 7b+tao+ ja is always a negative quantity.. Therefore, ža for |