1. Find two numbers such that the first with twice the second shall be 21, and twice the first, with half the second, shall be equal to 14. Ans. 5 and 8. 2. Find two numbers such that one half the first, with one third the second, may be 7, and one with one fourth of the second, may be 5.

third of the first,

Ans. 6 and 12.

added to the

3. Find two numbers such, that if 5 be first, the sum will be twice the second, and if four times the second be increased by 3, the sum will be three times the first. Ans. 13 and 9.

4. Find a fraction such, that if its numerator be increased by 1, and its denominator diminished by 2, its value will be, and if its numerator be increased by 3, and its denominator diminished by 2, the value will still be 3.

Ans. 22

5. Find a number, consisting of two digits, the sum of whose digits is equal to of the number, and the product of whose digits is equal to of the number. Ans. 36.

6. There is a certain number consisting of two figures, which is equal to 6 times the figure in the unit's place, and if 27 be added to the number, the digits will be inverted: what is that number? Ans. 36.

7. There is a certain number consisting of three figures, the sum of the digits is 7, twice the sum of the extreme digits is equal to 5 times the mean, and if 297 be subtracted from the number, the digits will be inverted: what is the number? Ans. 421.

8. Find three numbers, so that the first, with half the other two, the second with one-third of the other two, and the third with one-fourth of the. other two, may each be equal to 34. Ans. 10, 22, 26.

9. Find a number consisting of three figures, whose digits are in arithmetical progression, such that if this number be divided by the sum of its digits, the quotient will be 48; and if from the number 198 be subtracted, the digits will be inverted. Ans. 432.

10. If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C together in 10 days, in what time will each of them perform it alone? Ans. A in 14, B in 17, and C in 23

days.

QUADRATIC EQUATIONS.

80. QUADRATIC EQUATIONS may be divided into two kinds, namely, such as contain only the square of the unknown quantity, and those which contain both the square and first power of the unknown quantity; the first are called Pure Quadratics, and their solution may be effected by Rule V. of Simple Equations, with one unknown quantity; the second are called Adfected Quadratics, and they are of one of the four following forms:

ax2+ bx=+c,
ax2-bx=+c,
ax2+bx=-c,

ax2 bx-c.

All these are included in the general formula, ax2±bx =c; and we proceed now to solve this equation. If the first term were multiplied by 4a, it would become the square of 2ax; let both sides be multiplied by this quantity, viz. 4a, and the general equation becomes 4a2x2+4abx

4ac; the first side of this equation evidently wants something of being the square of a binomial, of which the first term is 2ax; let this quantity be p2, then adding this quantity to both sides, the equation becomes 4a2x2±1abx +p2=p2+4ac: now if the first side be a complete square, its square root can be no other than 2axp, (Art. 29 ;) hence squaring this, its square must be identical with the first side of the last equation, but its square is 4a2x2± 4apx+p2 to 4a2x2+4abx+p2; hence taking away the common terms from both sides, 4apx=4abx, and dividing both sides by 4ax, we have pb, and therefore the quantity which must be added to both sides of 4a2x2

4abx=4ac, so as to make it a complete square, is b2, that is, the square of the coefficient of x. If now the root of both sides be extracted, we obtain 2ax±b±±√b2±4ac, b±√ b2±4ac, which is a general

2a

formula in which, if we insert the proper values of a, b, and c, with their proper signs, (the sign of a being always +), we will have two values of x, both of which fulfil the conditions of the algebraic equation.

From the above investigation the following rule is de

rived :

RULE.

81. 1. Transpose all the terms containing the unknown quantity to one side of the equation, and so that the term containing the square of the unknown quantity may be positive, and the known terms to the other.

2. Multiply both sides of this equation by four times the coefficient of x2.

3. Add the square of the coefficient of x in the first equation to both sides, then will the first side be a complete square.

4. Extract the root of both sides, and the result will be a simple equation, which may be solved by the previous rules.

82. SCHOLIUM. When the equation, after being transposed as in the Rule, can be divided by the coefficient of x2, without introducing fractions, it may be solved conveniently as follows:-Perform the division, then add the square of half the new coefficient of x to both sides, which will make the first side a complete square; then extract the square root of both sides, and the equation will be reduced to a simple one, which may be solved as before.

EXAMPLE 1. Required the values of x in the equation 3x2-4x-160.

36x2-48x=1920,

by multiplying by 12=(4×3),

36x2-48x+16=1936, by adding (4) to both sides,

6x-4-44,

6x=4-44,

x=8, or -64,

by extracting the square root,
by transposition,

by dividing by 6.

Either of these values substituted in the given equation will make the sides equal, and they are therefore both roots of the equation.

EXAMPLE 2. Find two numbers whose sum is 100, and whose product is 2059.

Let the one; then since their sum is 100, the other may be represented by 100-x; and hence their product will be x(100-x), which by the question is equal to 2059. Hence 100x -x2=2059.

x2-100x=-2059, by changing the signs.
x2-100x+2500-441, by scholium.

x-50-21, by extracting the root.

... x=50±21=71, or 29, which are the two parts required, and therefore x has come out either the greater or the less part. The solution by the rule is left for the exercise of the pupil.

Ans. x-9, or -5.

1. Given x2=4x+45; to find x.
2. Given 5x2+7+4x=35; to find x.

Ans. x-2, or -23.

3. Given x2-50-5-6x; to find x.

Ans. x-5, or -11.

4. Given 4x-14= ; to find x. Ans. x=4, or -7.

14-x
x+1

7. Given x2+; to find x. Ans. x=1, or -. 8. Given 2x2+2x+6=30; to find x.

110

Ans.

x=3, or -4. 11, or -10.

9. Given x- -1= ; to find x. Ans.

10. Given =24x-700; to find x. Ans. x=70, or 50.

5

QUESTIONS PRODUCING QUADRATIC EQUATIONS.

1. What two numbers are those whose difference is 15, and half of whose product is equal to the cube of the less? Ans. 3 and 18.

whose product is 2059?

2. What two numbers are those whose sum is 100, and Ans. 71 and 29. 3. Find two numbers, so that their difference may be 8, and their product 240. Ans. 20 and 12. 4. Having sold a piece of cloth for L.24, I gained as much per cent. as the cloth cost me; what was its prime

cost?

Ans. L.20.

5. A grazier bought as many oxen as cost him L.480, and retaining 6 to himself, sold the remainder for the same sum, by which he gained L.4 a head on those sold. How many oxen did he buy, and what did he pay for each?

Ans. 30 oxen, at L.16 each. 6. A labourer dug two trenches, one of which was 4 yards longer than the other, for L.20, and each trench cost as many shillings a-yard as there were yards in its length. How many yards were in each? Ans. 12 yards and 16 yards. 7. The plate of a looking-glass is 24 inches by 16; it is

to be framed by a frame of uniform width throughout, whose surface shall be equal to the surface of the glass. Required the breadth of the frame. Ans. 4 inches. 8. There are three numbers in geometrical progression. The sum of the first and second is 10, and the difference of the second and third is 24.

What are the numbers?

Ans. 2, 8, 32. 9. 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 travel per hour?

Ans. A travelled 6 miles per hour, and B travelled 5. 10. A and B distribute L.1200 each among a certain number of persons. A relieves 40 persons more than B, and B gives L.5 a-piece to each more than A. How many persons were relieved by A and B respectively?

Ans. 120 by A, and 80 by B. 11. A person bought cloth for L.33, 15s., which he sold again at L.2, 8s. per piece, and gained as much by the bargain as one piece cost him. Required the number of pieces.

Ans. 15.

12. A company dine together at an inn for L.3, 15s. One of them was not allowed to pay, and the share of each of the rest was, in consequence, half-a-crown more than if all had paid. How many were in the company. Ans. 6.

13. A draper bought two pieces of cloth for L.3, 8s. The one was 6 yards longer than the other, and each of them cost as many shillings a-yard as there were yards in the piece. What was the length of each?

Ans. 2 yards and 8 yards. 14. Two girls carry 100 eggs to market. One of them had more than the other, but the sum which each received was the same. The first says to the second, if I had had as many eggs as you, I should have received 15 pence. The other answers, if I had had your number, I should have received 63 pence. How many eggs had

each, and what did each receive?

Ans. The first girl had 40, and the second 60, and each received 10 pence.

15. Find three numbers having equal differences, so that their sum may be 9, and the sum of their fourth powers 707. Ans. 1, 3, and 5. QUADRATIC EQUATIONS, WITH TWO UNKNOWN QUANTITIES.

83. In solving quadratic equations with two unknown quantities, it is necessary, frequently, to find a value of one