What are 4. The product of two numbers is 35, the difference of their cubes, is to the cube of their difference as 109 to 4. the numbers? Let x and y represent the numbers. Ans. 7 and 5. Then, by the conditions, xy=35, and x3-y3: (xy)3: 109:4 Divide by (xy) (Art. 42.) and a2+xy+y2: (x−y)2:: 109:4 Expanding, and x2+xy+y2 : x2—2xy+y2 :: 109:4 (Theorem 3.) 2 3xy: (x—y)2:: 105:4 But 3xy, we know from the first equation, is equal to 105. Therefore, (x-y)2=4, or x-y=2. We can obtain a very good solution of this problem by putting x+y=the greater, and x-y- the less of the two numbers. 5. What two numbers are those, whose difference is to their sum as 2 to 9, and whose sum is to their product as 18 to 77? Ans. 11 and 7. 6. Two numbers have such a relation to each other, that if 4 be added to each, they will be in proportion as 3 to 4; and if 4 be subtracted from each, they will be to each other as 1 to 4. What are the numbers? Ans. 5 and 8. 7. Divide the number 16 into two such parts that their product shall be to the sum of their squares as 15 to 34. Ans. 10, and 6. 8. In a mixture of rum and brandy, the difference between the quantities of each, is to the quantity of brandy, as 100 is to the number of gallons of rum; and the same difference is to the quantity of rum, as 4 to the number of gallons of brandy. How many gallons are there of each? Ans. 25 of rum, and 5 of brandy 9. There are two numbers whose product is 320; and the difference of their cubes, is to the cube of their difference, as 61 to 1. What are the numbers? Ans. 20 and 16. 10. Divide 60 into two such parts, that their product shall be to the sum of their squares as 2 to 5. Ans. 40 and 20. 11. There are two numbers which are to each other as 3 to 2. If 6 be added to the greater and subtracted from the less, the sum and the remainder will be to each other, as 3 to 1. What are the numbers? Ans. 24 and 16. 12. There are two numbers, which are to each other, as 16* to 9, and 24 is a mean proportional between them. What are the numbers ? Ans. 32 and 18. 13. The sum of two numbers is to their difference as 4 to 1, and the sum of their squares is to the greater as 102 to 5. What are the numbers? Ans. 15 and 9. 14. If the number 20 be divided into two parts, which are to each other in the duplicate ratio of 3 to 1, what number is a mean proportional between those parts? Ans. 18 and 2 are the parts, and 6 is the mean proportion between them. 15. There are two numbers in proportion of 3 to 2; and if 6 be added to the greater, and subtracted from the less, the results will be as 3 to 1. What are the numbers? Ans. 24 and 16. 16. There are three numbers in geometrical progression, the product of the first and second, is to the product of the second and third, as the first is to twice the second; and the sum of the first and third is 300. What are the numbers? Ans. 60, 120, and 240. 17. The sum of the cubes of two numbers, is to the difference of their cubes, as 559 to 127; and the multiplied by the second, is equal to 294. bers? square of the first, What are the num Ans. 7 and 6. 18. There are two numbers, the cube of the first is to the square of the second, as 3 to 1; and the cube of the second is to the square of the first as 96 to 1. What are the numbers? Ans. 12 and 24. SECTION VI. CHAPTER I. INVESTIGATION AND GENERAL APPLICATION OF THE BINOMIAL THEOREM. (Art. 127.) It may seem natural to continue right on to the higher order of equations, but in the resolution of some cases in cubics, we require the aid of the binomial theorem; it is therefore requisite to investigate that subject now. The just celebrity of this theorem, and its great utility in the higher branches of analysis, induce the author to give a general demonstration and the pupil cannot be urged too strongly to give it special attention. In (Art. 67.) we have expanded a binomial to several powers by actual multiplication, and in that case, derived a law for forming exponents and coefficients when the power was a whole positive number; but the great value and importance of the theorem arises from the fact that the general law drawn from that case is equally true, when the exponent is fractional or negative, and therefore it enables us to extract roots, as well as to expand powers. (Art. 128.) Preparatory to our investigation, we must prove the truth of the following theorem: If there be two series arising from different modes of expanding the same, or equal quantities, with a varying quantity having regular powers in each series; then the coefficients of the same powers of the varying quantity in the two series are equal. For example, suppose A+Bx+Cx2+Dx3, &c. =a+bx+cx2+dx3, &c. This equation is true by hypothesis, through all values of x. It is true then, when x=0. Make this supposition, and A=a. Now let these equal values be taken away, and the remainder divided by x. Then again, suppose x=0, and we shall find B=b. In the same manner we find C=c, D=d, E=e, &c. (Art. 129.) A binomial in the form of a+x may be put in the form х of *x(+); ; for we have only to perform the multiplication here indicated to obtain a+x. Hence, it will be sufficient to mul tiply every term of the expanded series by am for the expansion of (a+x), but as every power or root of 1 is 1, the first term х of the expansion of (1+2)" is 1, and this multiplied by a a m, what must give a for the first term of the expansion of (a+x)", ever m may be, positive or negative, whole or fractional. As we may put a in place of, we perceive that any bino a mial may be reduced to the form of (1+x), which, for greater facility, we shall operate upon. (Art. 130.) Let it be required to expand (1+x)", when mis a positive whole number. By actual multiplication, it can be shown, as in (Art. 67.) that the first term will be 1, and the second term mx. For if m=2, then If m=3, (1+x)TM==(1+x)2=1+2x, &c. And in general, (1+x)=1+mx+Ax2,+Bx3, &c. The exponent of x increasing by unity every term, and A, B, C, &c., unknown coefficients, which have some law of dependence on the exponent m, which it is the object of this investigation to discover. (Art. 131.) Now if m is supposed to be a fraction, or if m=the expansion of (1+x)m will be a root in place of a power, and 1 we must expand (1+x)”. For example, let us suppose r=2, then (1+x)=(1+x)3, and to examine the form the series would take, let us actually undertake to extract the square root of (1+x) by the common rule. Thus we perceive that in case of square root, the first term of the series must be unity, and the coefficient of the second term is the index of the binomial, and the powers of x increase by unity from term to term. We should find the same laws to govern the form of the series, if we attempted to extract cube, or any other root; but, to be general and scientific, we must return to the literal expression (1+x)". Now as any root of 1 is 1, the first term of this root must be 1, and the second term will have some coefficient to x. Let that coefficient be represented by p; and as the powers of x will increase by unity every term, we shall have 1 (1+x)=1+px+Ax2, &c. Take the r power of both members, and we shall have 1+x=(1+px+Ax2, &c.)r As r is a whole number, we can expand this second member by multiplication; that is, by (Art. 130.), the second member must take the following form 1+x=1+rpx+A'x2, &c. Drop 1, and divide by x, and we have 1=rp+A'x+, &c. |