Fibutes twice as much as a and 201. more ; and c as muck as A and B together. What sum did each contribute ? Aus: 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 ihen 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 many had each? Ans. A 22, B 26. 9 A person goes to a tavern with a certain quantiiy of money in his pocket, where he spends 2 shili'nigs; he then borrows as niuch 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. 38, 9d. 10. A man with his wife and child dine together at an inn. The landlord charged I shilling for the child; and for the woman he charged as much as for the child and 4 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 20d. 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 increas. ing the side by 1 man, he then wants 25 men to be a con.piete 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. 1. 14. The stock of three traders amounted to 8601. the shares of the first and second exceeded that of the third by by 240 ; and the sum of the 2d and 3d exceeded the first by 260. What was the share of each? Ans. The 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 78. and the whole reckoning 8 guineas. 17. A jocky has two horses : and also two saddles, the one valued at 187 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 Ist. What then were the values of the two horses? Ans. The 1st 61. and the 2d 91. 18. What two numbers are as 2 to 3, to each of which if 6 be added, thc 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, 3, 5 ? 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 ages? waken from the difference of their ages, the remainder will be equal to the age of the younger. What then were their Ans. 30 and 12. 24. To find four nunibers such, that the sum of the ist, 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 1st increased by 3, the second diminished by 3, the third mul:iplied 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 ax2 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 quanting in one term, and the first power in another term. As ax' + bor = 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. r$ + ax :6 2. x2 b 3. r2 - ax = -6. The generel niethod 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 ; nainely, by transposing all the terms which contain the unknown quantity to one side of the equation, and the known ternis 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 ihe solution. Then the proper solution is by completing the square as follows, viz. Von. R. Kk 2. Complete, 2. Complete the unknown side to a square, in this manner, viz. Take half the co-efficient of the second term, and square it ; whic:: 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 equations, 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 + n° is either tnor - n; for ++ x + n and - nx - # are each equal to + n2. But the square root of — 92, of ✓ -12, is imaginary or impossible, as neither + n nor - n, when squaredo gives 12, So, in the first form, x2 + ax = b, where x + La is found = ✓ 6 + fas, the root may be either + b + f62, or - Vo+*, since either of them being multiplied by itself produces b + tua, And this ambiguity is expre-sed by writing the uncertain or double sign + before vo + faz ; thus x = +6+ ful - fa. In this form, where x = Ivot tuz fa, the first value of x, viz, x= +6+412 - Ja, is always affirmative ; for since faz + 6 is greater than faz, the greater square must necessarily have the greater root; the:efore vot tas will always be greater than vlaz, or its equal ja ; and consequently + ✓ot 492 – fa wili always be affirmative. The second value, viz. x = vot uz ha will always be negative, because it is composed of two negative terms. Therefore when x2 + ax = b, we shall have x = + b + fa? fa for the affirmative value of x, and x =+ b + $a2 da for the nega. tive value of x. In the second form, where x = + b + tas + fa the first value, viz. x = + ✓bt fa3 + {a is always affirmative, since it is composed of two affirmative terms. But the second value, viz. ✓b + the + ja, will always be negative; for since 0 + \ais greater than la3, therefore 1 + $a2 will be greater than ✓ faz, or its equal fa; and consequently – Vot 1ns + la is always a negative quantity. Therefore, determined, making the root of the known side either + or -, which will give iwo rools of the equation, or two values of the unknown quantity. Note, 1. The root of the first side of the equation, is alwavs equal to the root of the first term, with half the co-efficient of the second term joined to it, with its sign, whether + or - 2. All equations, in which there are two terms, including the unknown quantity, and which have the index of the one just double that of the other, are resolved like quadratics, by completing the square, as above. Thus, * + ars = b, or xan + axb b, or x + ar c b, are the same as quadratics, and the value of the unknown quantity may be determined accordingly. Therefore, when xl - ax =b, we shall have x = tbt 1a2 + fa for tlic affirmative value of and ✓bt a2 + ta for the negat ve value of x; so that in both the first and becond forms, the unknown quantity has always two values, one of which is positive, and the other negative. But in the third form, where x = ✓ fa2 b + fa, both the values of x will be positive when 4a2 is gre::er than b. For the first value, siz, x= + 112 b + da will then be affirmative, being composed of two affirmative terms. The second value, viz. x = bt ha is affirma. tive also; for since fais greater than £a? – 6, therefore ✓ fund or fa is g eater than fal – b; and consequently ✓ 14% -6+ ja will always be an affirmative quan‘ity. So that, when x2 h, we shall have x = tv3.2 6 + ja, and also =6 + fa, for the values of x, both positive. But in this third form, if 6 be greater than faz, the solution of the proposed question will be impossible. For since the square of any quantity (whether that quantity be affirmative or negative) is always affirmative, the square root of a negative quantity is im. possible, and cannot be assigned But when b is greater than 4n2, then fa? - b is a negative quantity; and therefore its root v1- bis impossible, or imaginary ; consequently, in that case, hat v ta? — 6, or the two roots or values of xy are both impossible, or imaginary quantities, EXAMPLES |