4. Having sold a piece of cloth for 241., I gained as much per cent. as it cost me; what then was the price of the cloth?

Let x= pounds the cloth cost,

Then will 24-x= the whole gain,

But 100 x

24-x, by the question,

Or x2 = 100(24−x) = 2400–100x,

That is, x2 + 100x=2400,

Whence x=-50+√2500+ 2400 - 50+70=20, by the rule,

And consequently 201.=price of the cloth. 5. A person bought a number of sheep for 80l., and if he had bought 4 more for the same money, he would have paid 17. less for each; how many did he buy?

Let x represent the number of sheep,

x + 4

And 80x +320 = 80x + x2 + 4x, by the same, Or, by leaving out 8Or on each side, x2 + 4x=320, Whence x 2 +√4+320 −2+18, by the rule, And consequently x=16, the number of sheep. 6. It is required to find two numbers, such that their sum, product, and difference of their squares, shall be all equal.

Let x= the greater number, and y= the less.
Then { x + =
x+y=x_y} by the question,

Hence1=- =x-y, or x = y + 1, by 2d equation,

x + y

And (y+1)+y=y(y+1) by 1st equation, That is, 2y+1=y2+y; or y'-y=1, Whence y = + √(+1)='+'√5, by the rule,

1

3 1

And xy+1=2+ √5=2.6180&c.

=

7. It is required to find four numbers in arithmetical progression, such that the product of the two extremes shall be 45, and the product of the means 77.

Let x = least extreme, and y= common difference, Then x, x + y, x+2y, and x+3y, will be the four numbers,

Hence

{

x(x + 3y) = x2 + 3xy=45

by the x + y(x+2y) = x2 + 3xy + 2y2 = 77 Squest", And 2y=77-45 = 32, by subtraction,

32

Or y2=22=16 by division, and y=√16=4,

Therefore x2+3xy = x2 + 12x=45, by the 1st equat",

And consequently r = −6+√36 + 45,

x= −6+√36+ 45,=-6+9=3,

by the rule,

Whence the numbers are 3, 7, 11, and 15.

8. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84.

Let x, y, and z be the three numbers,
Then azy, by the nature of proportion,

Hence x+x=14-y, by the second equation,

And x2+2%x+x2=196-28y+y, by squaring

both sides,

Or x2+x2+2y=196-28+y by putting 2y for its equal 2xz,

That is x2+y+2=196-28y by subtraction,
Or 196-28y=84 by equality,

Hence y=

196-84
28

= 4, by transposition and div",

16

Again xx=y=16, or x= =10, by the 1st equation,

16

And x+y+x=- +4+14, by the 2d equation,

Or 16+4%+≈2=14%, or 22-10% -16,

Whence z = 5+√25 −16=5+3=8,0r3 by the rule, Therefore the three numbers are 2, 4, and 8.

9. It is required to find two numbers, such that their sum shall be 13(a), and the sum of their fourth powers 4721(b).

Let x= the difference of the two numbers sought,

But

(a + x) + (“ — x)* = b, by the question,

16

16

Or (a+x)*+ (α-x)=16b, by multiplication, Or 2a* +12a2x2 + 2x* = 16b, by involution and add", And x*+6a2x2=8b-a by transpos" and division, Whence r2=-3a2 + √9a* + 8b — a* = −3a2 + √8(a+b), by the rule,

And x=√3a2+2√2(a+b), by extracting the

root,

Where, by substituting 13 for a, and 4721 for b, we shall have x=3,

And 13-*= 10

=5, the less number,

2

2

10. Given the sum of two numbers equal to s, and their product equal to p, to find the sum of their squares, cubes, biquadrates, &c.

Let x and y denote the two numbers,

Ꮖ

Then will x+y=s, and xy=p, by the question. From the square of the first equation take twice the second, and we shall have

x2 + y2 = s2 - 2p, the sum of their squares. Multiply this by the 1st equation, and the product will be x3 + xy2 + x2y + y2 = s3 — 2ps.

From which subtract the product of the first and second equations, and there will remain

x13 + y3 = s3 − 3sp, the sum of the cubes. Multiply this likewise by the 1st equation, and the product will be x + xy+x3y + y* = s* — 3s2p. From which subtract the product of the second and third, and there will remain

x2 + y2 = s2 - 4s2p+2p, the sum of the biquadrates.

And, by proceeding to multiply the sum of the powers last found by the first equation, and subtracting from the result the product of the two next preceding equations, we shall obtain the sum of any powers proposed: thus,

x3 + y3 = s3 — 5s3p+5sp3,

the sum of the fifth powers, or, by expressing the theorem in general terms,

m(m-3)

2

2.3

2.3.4

'm(m—5)(m—6)(m −7)

Where it is evident that the series, or expression

for the sum of the powers, will terminate when the least index of s becomes =0.

QUESTIONS FOR PRACTICE.

1. It is required to divide the number 40 into two such parts, that the sum of their squares shall Ans. 23 and 17

be 818.

2. To find a number such, that if you subtract it from 10, and then multiply the remainder by the number itself, the product shall be 21.

Ans. 7 or 3 3. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their difference. Ans. 10 and 14

4. It is required to divide a line, of 20 inches in length, into two such parts, that the rectangle of the whole and one of the parts shall be equal to the square of the other. Ans. 10/5-10 5. It is required to divide the number 60 into two such parts, that their product shall be to the sum of their squares in the ratio of 2 to 5.

Ans. 20 and 40 6. It is required to divide 146 into two such parts, that the difference of their square roots may be 6. Ans. 25 and 121

7. What two numbers are those whose sum is Ans. 2 and 18

20 and their product 36?

8. The sum of two numbers is 14, and the sum of their reciprocals 3; required the numbers. Ans. and

9. The difference of two numbers is 15, and half their product is equal to the cube of the less number; required the numbers. Ans. 3 and 18

10. The difference of two numbers is 5, and the