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His loss per cent. on one pound is in

22

... his loss per cent. on x pounds must be 100, or r times as great. This gives the equation,

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x=50+ V100=50+10. Henee

x=60; x=40. Both these solutions equally fulfill the conditions of the problem.

Let us suppose, in the first place, that he paid 60 pounds for the horse; since he sold it for 24, his loss was 36. On the other hand, by the enunciation, his

60 X 60 loss was 60 per cent. on the original price; i. e., Too of thus 60 satisfies the conditions.

In the second place, let us suppose that he paid 40 pounds; his loss in this case was 16. On the other hand, his loss ought to be 40 per cent. on the

40

40 X 40 original price ; i. e., o of 40, or r=16; thus 40 also satisfies the conditions.

100

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100

GENERAL DISCUSSION OF THE EQUATION OF THE SECOND DEGREE. 191. The general form of the equation, the coefficients being considered independently of their signs, is

Xo+px+q=0.

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* In this and all the following values of r, calling the term? before the radical the rational part, and J +2 the radical part, we perceive that, when q is positive, the radical

part is greater than the rational, sinc alone equals, the rational part; and the sign of the whole expression is that of the radical part; but when q is negative, the radical part is less than the rational, and the sign of the whole expression is that of the rational part.

In this case, if we examine the general equation, we shall find that the conditions are absurd ; for, transposing 9, and completing the square, we have

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3

bat since - is, by hypothesis, a negative quantity, we may represent it by —m, where m is some positive quantity; then

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(+-+>) +m=0; that is, the sum of two quantities, each of which is essentially positive, is equal to 0, a manifest absurdity. Solving the equation,

a=FFV-m, and the symbol V-, which denotes absurdity, serves to distinguish this case. Hence, when the roots are imaginary, the problem to which the equation corresponds is absurd. We still say, however, that the equation has two roots ; for, subjecting these values of

to the same calculations as if they were real, that is, substituting them for x in the proposed equations, we shall find that they render the two members identical.

XVI. One case, attended with remarkable circumstances, still remains to be examined. Let us take the equation

ax'+bx-C=0.

-b+ + 4ac Whence

x= 20 Let us suppose that, in accordance with a particular hypothesis made on the given quantities in the equation, we have a=0; the expression for x then becomes

0

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The second of the above values is under the form of infinity, and may be considered as an answer, if the problem proposed be such as to admit of infinite solutions.

We must endeavor to interpret the meaning of the first,

In the first place, if we return to the equation ax2 + bx -c=0, we perceive that the hypothesis a=0 reduces it to br=c, whence we derive r=, a finite and determinate expression, which must be considered as representing the true value of in the case before us. That no doubt may remain on this subject, let us assume the equation

ax: + bx-C=0, and put x=, the expression will then become

a b

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Whence

cy: by-a=0. Let a=0, this last equation will become

cy-by=0,

b from which we have the two values y=0, y=-; substituting these values in

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* To show more distinctly how the indeterminate form arises, let us resume the general value of one of the roots.

__h+1 1+49c

If a were a factor of both the numerator and denominator, it might be suppressed, and then a, being put equal to zero, would give the true value of x. We can not, indeed, show the existence of this factor in the two terms of the fraction as it stands ; but if we multiply both numerator and denominator by — 6-1 12+ 4ac, it becomes

(-b+V 12+4ac)(-5--V 12+ 40C)

- 2a(6+1 be+tac)

With respect to the value r=", it is only to be observed that the divisor zero, having to be regarded as the limit of decreasing magnitudes, either positive or negative, it follows that the infinite value ought to have the ambiguous sign +. Thus the values of x, to recapitulate, become

r=1, r=+o. It is remarkable that, for this particular case, we have three values of x, while in the general case there are but two.

To comprehend how these values truly belong to the equation ara + b. -C=0, put it under the form

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When a=0, the question is to find values which will render

at zero.

We see that x=1 will do it; and as the same expression can be written under

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the form --t w e perceive that it becomes zero also, from the values r= #00.* XVII. Let us consider the still more particular case still, where we have,

0 at the same time, a=0, b=0. Then the two general values of x become We have seen above that the first may be changed into

20

it V 64+ 4uc Transforming the second in a similar manner, it becomes

(- - Vb2+ 4ac)(-6+7 +4ac) - 20 I=

201—6+ vb+4uc). -b+ 62+4ac In which, making a=1, b=0, the values of x, thus transformed, both give x=0; and here, also, the infinity ought to be taken with the sign +.

If we suppose a=0, b=0, c=0, the proposed equation will become altogether indeterminate.

The numerator, being the product of the sum and difference of two quantities, is equal to the dirference of their sqnares, io wit: 62(12+-4ac)=-4ac. We see, therefore, that 2a is a common factor to the numerator and denominator of the last expression. Suppressing it, we have

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* In the analytic theory of curves these values answer to the intersections of the axis of abscissas with the curve of the 3° order, the equation of which is yx-tbrf-c=0. If this curve be constructed, it will be found to cut the axis of abscissas first at a finite distance from the origin, and besides has this axis for an asymptote both on the side of the positive and negative abscissas, which amounts to saying that it cuts it at infinity in either die rection.

192. Let us now proceed to illustrate the principles established in this geceral discussion, by applying them to different problems.

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PROBLEM 5.
To find in a line, A B, which joins two lights of different intensities, a point
which is illuminated equally by each.
P,

A
À
P B

- P. (It is a principle in Optics that the intensities of the same light at different distances are inversely as the squares of the distances.)

Let a be the distance A B between the two lights.
Let b be the intensity of the light A at the distance of one foot from A.
Let c be the intensity of the light B at the distance of one foot from B.
Let P, be the point required.
Let A Pi=x; ... BP,=a-x.

By the optical principle above enunciated, since the intensity of A at the distance of 1 foot is b, its intensity at the distance of 2, 3, 4, ...... feet must be bbb. ã o ; hence the intensity of A at the distance of x feet must be . In the same manner, the intensity of B at the distance a—x must be a ng; but according to the conditions of the question, these two intensities are equal; hence we have for the equation of the problem

b c

72 (a—x)** Solving this equation, and reducing the result to its most simple form,

avb

x=Vbf We shall now proceed to discuss these two values : - avb

ave
1o. .......I=

- rbt vcl
Vo whence

- Votvo

- avc

a-r=76-Vc
I. Let 6>c.
- avb

lue of X, DILT is positive, and less than a, for is a proper fraction; hence this value gives for the point equally illuminated a point P,, situated between the points A and B. We perceive, moreover, that the point P, is nearer to B than to A; for, since b>c, we have

Vb 1
Votvo> Vb+ Vc, or 2 Vb> V6+ vc, and .. Vbi ve> ;

avba and, consequently, 7b+ VC

This is manifestly the result at which we ought to arrive, for we here suppose the intensity of A to be greater than that of B.

ave The corresponding value of a —X,

1, D IT is positive, and less than 2

- avb The second value of x, 7 I, is positive, and greater than a, for

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