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 loss was 60 per cent. on the original price; ¿. thus 60 satisfies the conditions. 60 100 of 60, or 60 × 60 -=36; 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 original price; i. e., ditions. 40 100 of 40, or 40 X 40 16; thus 40 also satisfies the con 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 * In this and all the following values of r, calling the term ±7 the radical part, we perceive that, when q is positive, the radical tional part, and part is greater than the rational, since √ alone equals the rational part; and the sign 4 2' 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 q, and completing the square, we have {XIII. p=0, x=±√q, the two values are equal with opposite signs. XIV. Let q be positive, {XIV. p=0, x=± √−q, both values are imaginary. XV. Let q=0, {XV. p=0, then r=0, or both values are equal to 0. p2 4 4 but since -q is, by hypothesis, a negative quantity, we may represent it by-m, where 4 that is, the sum of two quantities, each of which is essentially positive, is equal to 0, a manifest absurdity. Solving the equation, and the symbol √—m, 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 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 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, 0 с In the first place, if we return to the equation ar2+bx-c=0, we perceive that the hypothesis a=0 reduces it to br=c, whence we derive x= a finite and determinate expression, which must be considered as representing the true That no doubt may remain on this subject, let us assume the equation ax2+bx-c=0, b from which we have the two values y=0, y=; substituting these values in с To show more distinctly how the indeterminate form arises, let us resume the general value of one of the roots. 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 -b-b2+1ac, it becomes 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 r, to recapitulate, become 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 ax2+b.x -c=0, put it under the form When a=0, the question is to find values which will render will do it; and as the same expression can be written under we perceive that it becomes zero also, from the values XVII. Let us consider the still more particular case still, where we have, 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 Transforming the second in a similar manner, it becomes In which, making a=0, b=0, the values of x, thus transformed, both give x=; 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 difference of their squares, to wit: b2-(b2+4ac)=-4ac. We see, therefore, that 2a is a common factor to the numerator and denominator of the last expression. Suppress ing it, we have *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+b+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 beth on the side of the positive and negative abscissas, which amounts to saying that it cuts it at infinity in either direction. 192. Let us now proceed to illustrate the principles established in this gereral discussion, by applying them to different problems. 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 3 (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. 1 Let A P1=x; .. BP1=a-x. 1 ...... 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 bb b 4' 9' 16 ; hence the intensity of A at the distance of r feet must be b In the but с -x)2 same manner, the intensity of B at the distance a-r must be according to the conditions of the question, these two intensities are equal; hence we have for the equation of the problem Solving this equation, and reducing the result to its most simple form, a√b We shall now proceed to discuss these two values: is a proper fraction; hence this value gives for the point equally illuminated a point P1, 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 1 √b+√b> √b+ √c, or 2 √b> √b+ √c, and .. √b 1 ought to arrive, for we here suppose the intensity of A to be greater than that of B. The corresponding value of a-x, √ √e is positive, and less than a√b The second value of x, is positive, and greater than a, for - a |