I. Let b>c.

The first value of x, võ

αντ votvo

is positive, and less than a, for

is a proper fraction. Hence, this value gives for the point √b+ √ c illuminated equally, a point P situated between B and C. We perceive, also, that the point P is nearer to C than B; for, since b>c, we have võ+vb>võ+v«, or 2√/b>√b+ve, and αντ α 2

...

2, and, consequently,

This is manifestly correct, for the required point must be nearer the light of less intensity. The corresponding value of a-x, ανε √o+ve

is positive, and evidently less than

α

2'

This value gives a point P', situated on the prolongation of BC, and in the same direction from B as before. In fact, since the two lights emit rays in all directions, there will be a point P', to the right of C, and nearer the light of less intensity, which is illuminated equally by the two lights.

The second value of a-x, -ave

is negative, as it ought to b-ve

be, and represents the distance CP', in the opposite direction from C, (Art. 47.)

and positive. These values of x, and a-x, show that the point P is situated between B and C, but nearer to B than C. This is evidently a true result, since, under the present supposition, the intensity of the light B is less than that of the light C.

The second value of x,

ανδ

or

is essentially neg

ative, and represents a point P", in the opposite direction from B. As the intensity of the light B is now supposed to be less than that at C, there is, obviously, another such point of equal illumination.

The corresponding value of a—x, is

-avē

avē

It is

C

positive and greater than a, for vc>vc-võ

>1,

C

ave and vc-vb tances CB and BP", in the same direction from C as before.

>a. This represents CP", and is the sum of the dis

III. Let bc.

α

The first values of x and of a—x, reduce to which shows that

2'

the point illuminated equally is at the middle of the line BC, a result manifestly true, upon the supposition that the intensities of the two lights are equal.

The other two values are reduced to

αντ

0

(Art. 136.) This result is manifestly true, for the intensities of the two lights being supposed equal, there is no point at any finite distance, except the point P, which is equally illuminated by both.

IV. Let bc, and a=0.

The first system of values of x and a-x, become 0. This is evidently correct, for when the distance BC becomes 0, the distances BP and CP also become 0.

0

The second system of values of x and a-x, become symbol of indetermination, (Art. 137.)

this is the

This result is also correct, for if the two lights are equal, and placed at the same point, every point on either side of them will be illuminated equally by each.

All the values of x and a- reduce to 0; hence, there is no point equally illuminated by each. In other words, the solution of the problem fails in this case, as it evidently should.

This might also have been inferred from the original equation;

1. Required a number such, that twice its square, increased by 8 times the number itself, shall be 90.

How may the question be changed, that the negative answer, taken positively, shall be correct in an arithmetical sense?

2. The difference of two numbers is 4, and their product 21. Required the numbers.

Ans. +3, +7, or 3 and 7. 3. A man bought a watch, which he afterward sold for $16. His loss per cent. on the first cost of the watch, was the same as the number of $'s which he paid for it. What did he pay for the watch? Ans. $20, or $80.

4. Required a number such, that the square of the number increased by 6 times the number, and this sum, increased by 7, the result shall be 2. Ans. x——1, or -5.

What do the values of x show? How may the question be changed to be possible in an arithmetical sense?

5. Divide the number 10 into two such parts, that the product shall be 24. Ans. 4 and 6, or 6 and 4.

Is there more than one solution? Why?

6. Divide the number 10 into two such parts that the product shall be 26. Ans. 5+√—1, and 5—√—1.

What do these results show?

7. The mass of the earth is 80 times that of the moon, and their mean distance asunder 240000 miles. The attraction of gravitation being directly as the quantity of matter, and inversely as the square of the distance from the center of attraction, it is required to find at what point on the line passing through the centers of these bodies, the forces of attraction are equal.

Ans. 215865.5+ miles from the earth,

and 24134.5— ""

Or, 270210.4+

แ

66 moon.

"earth,

This question involves the same principles as the Problem of the Lights, and may be discussed in a similar manner. The required results, however, may be obtained directly from the values of x, page 208, calling a=240000, b=80, and c=1.

て

T

vie

TRINOMIAL EQUATIONS.

240. A Trinomial Equation is one consisting of three terms, the general form of which is axbx"c.

Every trinomial equation of the form

that is, every equation of three terms containing only two powers of the unknown quantity, and in which one of the exponents is double the other, can be solved in the same manner as an affected equation.

As an example, let it be required to find the value of in the equation

x*—2px2-q.

Completing the square, x-2px2+p2=q+p2.

x2—p=√1+p2.

x2=+p±vI+P2.

241. Binomial Surds. -Expressions of the form A±√B, like the value of x2 just found, or of the form √A±√В, are called Binomial Surds.

The first of these forms, viz., A±√B, frequently results from the solution of trinomial equations of the fourth degree; and as it is sometimes possible to reduce it to a more simple form by extracting the square root, it is necessary to consider the subject here.

We shall first show that it is sometimes possible to extract the square root of A±√B, or to find the value of A±√/B.

Let us inquire how such binomial surds may arise from involution.

If we square 2±√/3, we have 4±4√/3+3, which, by reduction, becomes 7±4/3. Hence, 7±4/3=2±√3. In the same way V

it may be shown that

√√5±2√/6=√2±√3.

In

It thus appears that the form A±√В may sometimes result from squaring a binomial of the form a±√/b, or √a±√6, and uniting the extreme terms, which are necessarily rational, into one. such cases, A is the sum of the squares of the two terms of the root, and B is twice their product.

To find the root, therefore, put x2+y2=A and 2xy=√B, and proceed to find x and y, the terms of the root. Thus,

Subtracting, we have x2-2xy+y2-7—4√/3.