{XIIL p = 0, XII. p negative, x = +/+, one value =+p, the other = 0. XIIL Let q be negative, XIV. Let q be positive, q, the two values equal with opposite signs 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 z 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 endeavour 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 bx + c, whence we derive с x = a finite and determinate expression, which must be considered as b representing the true value of in the case before us. That no doubt may remain on this subject, let us assume the equation If. in addition to the hypothesis a=0, we have also b= 0, the value In fact, the equation cy2 — by — a = 0), under this double hypothesis, is reduced to cy2 : = 0, an equation in which the values of y are equal, and each = 0. Hence the two corresponding values of x will both be infinite. If we suppose a = 0, b = 0, c = 0, the proposed equation will become altogether indeterminate. 170. Let us now proceed to illustrate the principles established in this general 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 A PB 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, ... ... b be the intensity of the light A at the distance of one foot from A, c be the intensity of the light B at the distance of one foot from B, P1 be the point required, By the optical principle above enunciated, since the intensity of A at the distance of 1 foot is h, 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 x feet must be In the same manner, the intensity of B at the distance a— 2 must be (a с x )2 : but according to the conditions of the question, these two intensities are equal, hence we have for the equation of the problem, Bolving this equation, and reducing the result to its most simple form, T = ab 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 P1 is nearer to B than to A; for, since bc, we have, 7 vo + √o > vb + √c, or, 2 √b> √b + √c, and .. and consequently, a b a 2* 1 √b 7 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. avc ર √b+√e is positive, and less than - cond value gives a point P2, situated in the production of A B, and to the right of the two lights. In fact, we suppose that the two lights give forth rays in all directions, there may therefore be a point in the production of A B equally illuminated by each, but this point must be situated in the production of A B to the right, in order that it may be nearer to the less powerful of the two lights. It is easy to perceive why the two values thus obtained are connected by the same equation. If, instead of assuming A P1 for the unknown quantity x, we take A P2, then B P1 = x- a, thus we have the equation 2 -- 2 (x- since (x — a)2 is identical with (a — x)2, the new equation is the same as that already established, and which consequently ought to give A P2 as well as A P1. a; but changing the signs of the equation a — x = of B P2. we find and this value of x— a represents the absolute length II. Let bc. avb The first value of x, btve is positive, and less than >√o+ √b.: 16+ √072 0. The corresponding value of a — x, avc is positive, and greater than Hence, the point P1 is situated between the points A and B, and is nearer to A than to B. This is manifestly the true result, for the present hypothesis supposes that the intensity of B is greater than the intensity of A. 16 order to interpret the signification of this result, let us resume the original b 2 x с ·x)' equation and substitute -x for x, it thus becomes = +2)■• But since (ax) expresses in the first instance the distance of B from the point required, a+x ought still to express this same distance, and therefore the point required must be situated to the left of A, in P3, for example. In fact, since the intensity of the light B, is, under the present hypothesis, greater than the intensity of A, the point required must be nearer to A than to B. The corresponding value of a- -x, a or a√c is positive, and the reason of this is, that, z being negative, a―x expresses, in reality, an arithmetical sum. III. Let b = c. The first two values of x and of a-x are reduced to which gives the a 2' bisection of A B for the point equally illuminated by each light, a result which is manifestly true, upon the supposition that the intensity of the two lights is the same. a The other two values are reduced to that is, they become infinite, that is to say, the second point equally illuminated is situated at a distance from the points A and B greater than any which can be assigned. This result perfectly corresponds with the present hypothesis; for if we suppose the difference b—c, without vanishing altogether, to be exceedingly small, the second point equally illuminated, exists, but at a great distance from the two lights; this is indicated the denominator of which is exceedingly small in comparison with the numerator if we suppose b very nearly equal to c. In the extreme case, when b⇒c, or √✓/b-√c, the point required no longer exists, or is situated at an infinite distance. IV. Let bc, and a = 0. The first system of values of x and a―x in this case become 0, and the second system. This last result is here the symbol of indetermination; for if we recur to the equation of the problem an equation which can be satisfied by the substitution of any number whatever for 2. In fact, since the two lights are supposed to be equal in intensity and to be placed at the same point, they must illuminate every point in the line A B equally. The solution 0, given by the first system, is one of those solutions, infinite in number, of which we have just spoken. V. Let a =0, b not being = c. Each of the two systems in this case is reduced to 0, which proves, that in this case, there is only one point equally illuminated, viz. the point in which the two lights are placed. The above discussion affords an example of the precision with which algebra answers to all the circumstances included in the enunciation of a problem. We shall conclude this subject by solving one or two problems which require the introduction of more than one unknown quantity. Problem 6 To find two numbers such that when multiplied by the numbers a and b respectively, the sum of the products may be equal to 2 s, and the product of the two numbers equal to p. be Let x and y be the two numbers sougnt, the equations of the problem will Substituting this value in (2) and reducing, we have a x2 - 2 s x + bp = 0 The problem is, we perceive, susceptible of two direct solutions, for s is manfestly >√s2 —a bp; but in order that these solutions may be real we must have s2 or a bp. Let a = = 1; in this case the values of x and y are reduced to |