The sum of two numbers is 35 and their product is 15 find the sum of their reciprocal

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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, =

= 45

Hence {(x + y) (x+2y) = x2 + 3xy + 2y2 = 77

question,

And 2y77-45 32, by subtraction,

Or, y2 = = 16 by division, and y = √16 = 4. 2

Therefore, a+ 3xy = x2+12x 45, by the 1st equation, And consequently, 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, xz y, by the nature of proportion, x + y + z = = 14

And

484 by the question,

Hence, a 14-y, by the second equation, + z = And x2+2zx + z2: = 196 28y + y2, by squaring both sides, Or x2+22+2y= 196 28y+y2 by putting 2y for its equal 2xz,

y2 z2 =

That is, a+ya+22 19628y, by subtraction, Or, 19628y = 84, by equality,

196 84

Again, xzy2 = 16, or x =

4, by transposition and division.

, by the 1st equation,

And x + y + z =

+4+ z = 14, by the 2d equation,

Or, 16+4z + z2. 142, or 22-10z = -16, Whence z= 5±√(2516) 5±38, or 2 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). the difference of the two numbers sought, 1

Let x= a + x Then willa+x, or

a-x

1 a- X, 2

(a + x), (α — x)a

But

Or (a + x)+(ax)

16b, by multiplication,

= =

Or 2a+12a2x2+2x=166, by involution and addition, And 4+6a2x2 8b a, by transposition and division, Whence x2 - ―― 3a2 + √ (9a1 + 8b — a1) = 3a2 + √ 8 (a + b), by the rule,

And x = v[-3a2 + 2 √ 2 (a* +6)], by extracting the root, Where, by substituting, 13 for a, and 4721 for b,

we shall have x = 3,

13 -X

And

5, the less number.

2 The sum of which is 13, and 84 + 5 = 4721.

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

Let x and y denote the two numbers; then

(1.) x + y = s, (2.) xy = p.

From the square of the first of these equations take twice the second, and we shall have

(3.) x2 + y2 s2 - 2p sum of the squares. Multiply this by the 1st equation, and the product will be x3 + xy2 + x3y + y3 = s3 — 2sp. From which subtract the product of the first and second equations, and there will remain (4.) x3 + y3 = s3 . 3sp= = sum of the cubes. Multiply this likewise by the 1st, and the product will be x2+xy3 + x3y + y = s*- 3s2p; from which subtract the product of the second and third equations, and there will remain

(5.) x2 + y2 = s1 — 4s2p + 2p2 = sum of the biquadrates. And, again multiplying this by the 1st equation and subtracting from the result the product of the second and fourth, we shall have

(6.) x3 + y5 -- s5 5s3p+5sp2= sum of the fifth powers. And so on; the expression for the sum of any powers in general being a + ym = sm — msm-2p+ m(m−4)(m−5)

m(m-3)

2.3

m(m — 5)(m — 6) (m — 7)

2·3:4

&c. Where it is evident that the series will terminate when the index of s becomes = 0.

1. It is required to divide the number 40 into two such parts, that the sum of their squares shall be 818.

Ans. 23 and 17.

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 shall 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.

in the ratio of 2 to 5.

Ans. 105 10, and 30 — 10 √5. 5. It is required to divide the number 60 into two such parts, that their product shall be to the sum of their squares Ans. 20 and 40. 6. It is required to divide the number 146 into two such parts, that the difference of their square roots shall be 6. Ans. 25 and 121. 7. What two numbers are those whose sum is 20 and their product 36? Ans. 2 and 18.

8. The sum of two numbers is 11, and the sum of their reciprocal 3; required the numbers. Ans. and 5. 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 difference of their cubes 1685; required the numbers. Ans. 8 and 13.

11. A person bought cloth for 337. 15s., which he sold again at 21. 8s. per piece, and gained by the bargain as much as one piece cost him; required the number of pieces.

Ans. 15.

12. What two numbers are those, whose sum, multiplied by the greater, is equal to 77, and whose difference, multiplied by the less, is equal to 12? Ans. 4 and 7.

13. A grazier bought as many sheep as cost him 607., and after reserving 15 out of the number, sold the remainder for 541., and gained 2s. a head by them: how many sheep did he buy? Ans. 75. 14. It is required to find two numbers, such that their

product shall be equal to the difference of their squares, and the sum of their squares equal to the difference of their cubes. Ans. 1 ✓5 and (5+ √5). 15. The difference of two numbers is 8, and the difference of their fourth powers is 14560; required the numbers.* Ans. 3 and 11.

16. A company at a tavern had 81. 15s. to pay for their reckoning; but before the bill was settled, two of them went away; in consequence of which those who remained had 10s. apiece more to pay than before; how many were there in company?

Ans. 7.

17. A person ordered 71. 4s. to be distributed among some poor people; but before the money was divided, there came in, unexpectedly, two claimants more, by which means the former received a shilling apiece less than they would otherwise have done; what was their number at first?

Ans. 16 persons. 18. It is required to find four numbers in geometrical progression such, that their sum shall be 15, and the sum of their squares 85. Ans. 1, 2, 4, and 8. 19. The sum of two numbers is 11, and the sum of their fifth powers is 17831; required the numbers. Ans. 4 and 7. 20. It is required to find four numbers in arithmetical pro

* In solving this question, the reduced equation, found by the usual methods of operation, will be of the form x3+ax = b; which is a cubic equation, and therefore cannot be resolved by the ordinary rules of quadraties; but such equations may sometimes be reduced to the form of a quadratic, and then resolved according to the rules already given. Whenever, in a cubic equation of the form x3 + ax = b, b can be divided into two factors m and n, so that mean, then the cubic equation can be resolved as a quadratic: thus, in the cubic equation x36x=20, 20 = 2 × 10, and 22+6= 10. Now, multiplying both the sides of the equation by x, we have x4+6x2 = 10 × 2x, adding (2x)2 to both sides, x + 10.x2 = (2x)2+10 (2x); completing the square, x1+10.x2+25= (2x)2 + 10 (2x) +25,

and extracting the root, x2+5=2x+5; .. by transposition, x2=2x, and x2, or = 0.

This method, as well as some other similar artifices, is of no utility when the divisor has not integral roots, and even then it can be resolved more readily by Newton's Method of Divisors.

It is proper to observe, that cubic equations of the form x3+a+ bec, may be also exhibited under the form of a quadratic, from which, by completing the square, the value of the unknown quantity will be determined. For instance, the cubic equation x3+2ax2+5a2x+4a3 =0, may be reduced to the form (x2+ax)2+4a2 (x2+ax)=0; thus, multiply the given equation by x, we have x4 +2a.x35a2x2-4a3· x=0; which may be readily exhibited under the above form. See Ryan's Elementary Treatise on Algebra, Practical and Theoretical. (Art. 423.).-ED.

gression, such, that their common difference shall be 4, and their continued product 176985. Ans. 15, 19, 23, and 27. 21. Two detachments of foot being ordered to a station at the distance of 39 miles from their present quarters, begin their march at the same time; but one party, by travelling of a mile an hour faster than the other, arrive there an hour sooner; required their rates of marching.

Ans. 31 and 3 miles per hour. 22. It is required to find two numbers, such, that the square of the first plus their product shall be 140, and the square of the second minus their product 78. Ans. 7 and 13.

2167

23. It is required to find two numbers, such that their difference shall be 13,5%, and the difference of their cube roots 11. Ans. 155, and 21. 24. It is required to find three numbers in arithmetical progression, such that the sum of their squares shall be 93; and if the first be multiplied by 3, the second by 4, and the third by 5, the sum of the products shall be 66. Ans. 2, 5, and 8.

25. The sum of three numbers in harmonical proportion is 191, and the product of the first and third is 4032; required their numbers. Ans. 72, 63, and 56. 26. It is required to find four numbers in arithmetical progression, such that if they are increased by 2, 4, 8, and 15 respectively, the sums shall be in geometrical progression. Ans. 6, 8, 10, and 12. 27. It is required to find two numbers, such, that if their difference be multiplied into their sum, the product will be 5; but if the difference of their squares be multiplied into the sum of their squares, the product will be 65. Ans. 3 and 2. 28. It is required to divide the number 10 into two such parts, that if the square root of the greater part be taken from the greater part, the remainder shall be equal to the square root of the less part added to the less part.

Ans. 5+ 19 and 5—19.

29. It is required to find two numbers, such that if their product be added to their sum, it shall make 61; and if their sum be taken from the sum of their squares, it shall leave 88. Ans. 72 and 7 v 2. 30. It is required to find two numbers, such that their difference multiplied by the difference of their squares, shall be 576; and their sum multiplied by the sum of their squares, shall be 2336. Ans. 5 and 11.

31. It is required to find three numbers in continual proportion, whose sum shall be 20, and the sum of their squares 140. Ans. 63, 63, and 63 - √ 31.