Let x least extreme, and y= common difference, Then x, x+y, x+2y, and x+3y, will be the four numbers, Hence {(x+y)(x+2y)=x2 + 3xy + 2y2 = 77 question, x(x+3y)=x2+3xy=45 77} by the And 2y277-45-32, by subtraction, 32 Or y2=16 by division, and y=√16=4, 2 Therefore x+Sxy=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. And { } by the question, Let x, y, and z be the three numbers, Then xzy, by the nature of proportion, Sx+y+x=14 { x2 + y2+z2=84by the question, Hence x+14-y, by the second equation, x2+2zx+z2196-28y+y, by squaring both sides, And Or x2+z3+2y3=196 −28y+y2 by putting 2y2 for its equal 2xz, That is x+y+22=196 28y by subtraction, Hence y= 196-84 28 =4, by transposition and division, 16 Again xz=y=16, or x, by the 1st. equation, 16 2 And x+y+z+4+2=14, by the 2d equation, 2 Or 16+42+z2=142, or 23 .10z=-16, Whence 25±√(25—16)=5±3=-8, 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). Let x= the difference of the two numbers sought, Then willa+x, 1 1 a+x or 2 the greater number, And-a- or == the less, 2 (a+x), (α—x) 4 16 16 -=b, by the question, Or (a+x)+(a-x)=16b, by multiplication, Or 2a+12a3x2+2x4166, by involution and addition, And x+6ax=8b-a4, by transposition and division, Whence x2=-3a2+ √ (9a1+8b—a1) —— 3α2 = 8(a+b), by the rule, 4 And x-3a2+22(a+b), by extracting the root, Where, by substituting 13 for a, and 4721 for b. we shall have x=3, 13+x 16 Therefore =8, the greater number. 5, the less number, The sum of which is 13, and 84+54=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 x2+xy2+x2y+y3=s3-2sp. From which subtract the product of the first and second equations, and there will remain (4.) x3 y3s3-3sp-sum of the cubes. Multiply this likewise by the 1st, and the product will be x2+xy3+x2y+ya=sa—3x2p ; from which subtract the product of the second and third equations, and there will remain (5.) x+y=s—4s3p+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.) x5+5=55 - 5s3p+5sp2= sum of the fifth powers. And so on; the expression for the sum of any powers in m(m-3) general being am+y = sm-msm-2p+- 2sm4pa—— 2 m(m—5)'m—6) (m—7) 2.3.4 Where it is evident that the series will terminate when the index of s becomes = o. QUESTIONS FOR PRACTICE. 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. 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 in the ratio of 2 to 5. Ans. 20 and 40 6. It is required to divide the number 146 into such two 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 14, and the sum of their reciprocals 31; 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 difference of their cubes 1685; required the numbers. Ans. 8 and 18 11. A person bought cloth for 331. 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 60l., and after reserving 15 out of the number, sold the remainder for 547., 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. ✓5 and 1(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 *In solving this question, the reduced equation, found by the usual me⚫ thods 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 quadratics; 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+axb; b can be divided into two factors m and n, so that m2 fan, then the cubic equation can be resolved as a quadratic: thus, in the cubic equation x3+6x=20, 202X 10, and 22 +6 10. Now, multiplying both the sides of the equation by x, we have x4+6x2 = 10X2x, adding (2x)2 to both sides, x4 +10x2 (2x)2 + 10(2x); .. completing the square, x4+10x225 (2x)2 + 10(2x) + 25, and extracting the root, x2 +5=2x+5;.. by transposition, x2 = 2x, and x=2, or =0. This method as well as some other similar artifices, is of no utility when the equation has not integral roots, and even then it can be resolved more readi ly by Newton's Method of Divisors. It is proper to observe that cubic equations of the form x3 +ax 3+ bx=c, may be also exhibited under the form of a quadratic, from which by complet 0 16 A company at a tavern had 87. 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. a piece 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 a piece 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 progression 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. 23. It is required to find two numbers, such that their difference shall be 13, and the difference of their cube roots 1. Ans. 15, and 21. ing the square, the value of the unknown quantity will be determined. For instance, the cubic equation x3+2ax2+5x2x+4a3 =0, may be reduced to the form (3+ax) 3+4a2 (x2+ax)=0; thus, multiply the given equation by x, we have x4+2ax3+5a2 x2 +4a3x=0; which may be readi. ly exhibited under the above forin: see Ryan's Elementary Treatise on Algebra, Practical and Theoretical. (Art. 423.) ED. |