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present, in the ratio of 7 to 5, but that 30 years ago, they were in the ratio of 2 to 1; what are their ages? Ans. 70 and 50.

30. A. and B. began trade with equal sums of money. In the first year A. gained £40, and B. lost £40; but in the second, A. lost one third of what he then had, and B. gained a sum which was £40 less than twice what A. had lost; when it appeared, that B. had twice as much as A.; what sum had each of them at first? Ans. £320.

31. A farmer sold 96 loads of hay to two persons. To the first one half, and to the second one fourth of what his stack contained; how many loads were in the stack?

Ans. 128.

32. If 116 be divided into four parts in such manner, that the first being increased by 5, the second diminished by 4, the third multiplied by 3, and the fourth divided by 2, the results will all be the same; what are the parts? Ans. 22, 31, 9, and 54.

33. A shepherd in time of war, was plundered by a party, who took from him of his flock, and of a sheep more; another'party took from him 4 of what he had left, and of a sheep more; afterwards, a third party took. of what remained, and a sheep more, when he had but 25 left; how many had he at first? Ans. 103.

34. A person has two horses and a gig; the gig is worth 150 dollars. When the first horse is attached to the gig, the value of the two is twice that of the second; but when the second horse is put to the gig, the value is three times that of the first horse; what were the horses respectively worth?

Ans. the first 90 dollars, the second 120. 35. When a company at a tavern came to pay their reckoning, they found, that if there had been three persons more, they would have had a shilling a piece less to pay; but if there had been two less, they would have had a shilling a piece more to pay; required the number of persons, and the quota of each?

Ans. 12 persons, quota of each 5s.

36. There are three equidifferent numbers, whose sum is 324, and the first is to the third as 5 to 7 ; what are the numbers? Ans. 90, 108, and 126.

37. A man and his wife usually drank a cask of beer in 12 days; but when the man was from home, it lasted the woman 30 days; how long would the man alone be in drinking it? Ans. 20 days.

38. A. and B. can perform a piece of work in 8 days, A. and C. in 9 days, and B. and C. in 10 days; how many days would they severally require to perform the same work ? Ans. A. 1434 days, B. 1723, and C. 2331. 39. The hypothenuse of a right angled triangle is 13, and the area 30; what are the other two sides ?*

Ans. 12 and 5.

40. There are four numbers in geometrical progression; the sum of the extremes is 18, and the sum of the means 12; what are the numbers? Ans. 2, 4, 8, and 16.

41. Required to find four numbers in geometrical progression, such that the difference of the means shall be 100, and the difference of the extremes 620.

The numbers are 5, 25, 125, and 625.

SECTION VII.

QUADRATIC EQUATIONS.

60. When an equation contains the square and simple power of an unknown quantity; or in general, two powers, one of whose indices is double the other, of an unknown quantity, whether that unknown quantity be simple or compound, it is called an adfected quadratic equation.

To solve this problem, it must be recollected that the square of the hypothenuse is equal to the sum of the squares of the other two sides, and that the area is half the product of those sides.

RULE 1.

Arrange the terms, by transposition or otherwise, so that the highest power of the unknown quantity shall be contained in the first, and the other power in the second term on the left side of the equation; and the term or terms consisting of known quantities, on the other side.

When the highest power of the unknown quantity contains a co-efficient, divide the equation by that coefficient. Then add, to each member of the equation, the square of half the co-efficient of the unknown quantity, (or the numerical part of the second term,) and the first member of the equation will be a complete square.

Extract the root and find the value of the unknown quantity by the former methods.

RULE 2.

61. Arrange the terms as before. When the highest power of the unknown quantity has a co-efficient, multiply the equation by four times that co-efficient, and to each member of the resulting equation, add the square of the co-efficient of the unknown quantity, contained in the primary equation; then the first member of the resulting equation will be a complete square.

Extract the root, and proceed by the former methods, to determine the value of the quantity sought.

EXAMPLES.

Given 2x2-9x=266, to find the value of x.

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By rule 2.-Multiplying by 8, 16x3-72x=2128.

Adding 81 to each member, 16x2-72x+81=2209. Extracting the square root, 4x-9=47.

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2. Given ax+bx-c, to find the value of x.

By rule 1.-Dividing by a, x2+

to each, x2+x+

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b

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α 4a2

4a2

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√b2+4ac-b

By transposition, x= 2α

By rule 2.-Multiplying by 4a, 4a2x2+4abx=4ac. Adding b2 to each, 4a2x2+4abx+b2=4ac+b2.

By evolution, 2ax+b=4ac+b2.

By transposition and division, x=

√4ac+b2-b

2

2a

3. Given 2x3+3x=2, to find the value of x.

By rule 1. x3+x3=1.

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By rule 2. 16x3+24x3—16.

Adding 9 to each, 16x3+24x3+9=25.

By evolution, 4x+3=5.

By transp. and division, x3- 1-4

Therefore, x===

4. Given x1-12x3+44x2-48x=9009, to determine 2. This equation is equivalent to the following:

x-12x+36x+8x48x=9009,

Or to this (x2-6x)2+8.(x2-6x)=9009.

In which, x2-6x may be considered as the unknown quantity.

Hence, proceeding as in the former examples,

(x2-6x)2+8.(x2-6x)+16=9025.

.. x2-6x+4=95, or x2-6x=91.

Whence, x2-6x+9=100.

..x-3=10, or x=13.

5. Given √(x———-_)+√(1———-1)=x, =x, to find the value of x.

Multiplying by √x, √x2−1+√x—1=x√x.

G

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