Changing the means x+y: x-y:: 42: 6

2. Divide the number 14 into two such parts, that the quotient of the greater divided by the less, shall be to the quotient of the less divided by the greater, as 16 to 9.

Let x= the greater part, and 14-x= the less.

3. There are three numbers in geometrical progression whose sum is 52, and the sum of the extremes is to the mean as 10 to 3. What are the numbers? Ans. 4, 12, and 36. Let x, xy, xy2 represent the numbers.

Then, by the conditions, x+xy+xy2=52 (1)

52

52

From equation (1), x=

1+y+y2 1+3+9

52 =4. 13

4. The product of two numbers is 35, the difference of their cubes, is to the cube of their difference as 109 to 4. What are the numbers? Ans. 7 and 5.

Let x and y represent the numbers.

Then, by the conditions, xy=35, and x3-y3: (x-3) :: 109: 4
Divide by (x-y) (Art. 42.) and x2+xy+y2: (x-y)2 :: 109: 4
Expanding, and
(Theorem 3.)

x2+xy+y2: x2-2xy+y2 :: 109:4

Зху: (х-у)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. their product shall Ans. 40 and 20.

10. Divide 60 into two such parts, that be to the sum of their squares as 2 to 5. 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 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 their product as 34 to 15. 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.

SECTION VI.

CHAPTER I.

L

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 like 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 r. 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. &c.

(Art. 129.) A binomial in the form of a-x may be put in the form of ax ; for we have only to perform

ax(1+);

the multiplication here indicated to obtain a+x. Hence,

to multiply every term of the expanded series by am for the expansion of (a+x)m, but as every power or root of 1 is 1, the first term of the expansion of (a+x)m must be am, whatever m may be, whole or fractional.

As we may put z in place of ,we perceive that any binomial 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 m is 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

(1+x)=(1+x)2=1+2x, &c.

If m=3, (1+x)=1+3x, &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 is the object of this investigation to discover.

(Art. 131.) Now if m is supposed to be a fraction, or if m=1, the expansion of (1+x)m will be a root in place of a power, and we must expand (1+x).

r

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. That is, (1+x)=1+px+Ax2, &c. Take the r power of both members, as in (Art. 130.) and

1+x=(1+px+ &c.)=1+rpx+Ax2, &c. Drop 1, and divide by x, and 1=rp+Ax+Bx2, &c. This equation is true for all values of x; therefore, by (Art. 128.) p=1; hence (1+x)=1++x+Ax2+Bx3 &c.; the same general form as when the exponent was considered as a whole number.