In the first and second forms, the quantity under the radical sign will be positive, whatever be the relative values of Ρ and q, since q and p2 are both positive; and therefore, both roots will be real. And since q + p2 > p2, it follows that, √1 + p2 >p, and consequently, the roots in both these forms will have the same signs as the radicals. In the first form, the first root will be positive and the second negative, the negative root being numerically the greater. In the second form, the first root is positive and the second negative, the positive root being numerically the greater In the third and fourth forms, if they will have the same sign as the entire part of the root Hence, both roots will be negative in the third form, and both positive in the fourth. p2 = If p2q, the quantity under the radical sign becomes 0, and the two values of x in both the third and fourth forms will be equal to each other; both equal to -p in the third form, and both equal to +p in the fourth. If p2, the quantity under the radical sign is negative, and all the roots in the third and fourth forms are imaginary. But from the third principle demonstrated in Art. 116, the greatest value of the product of the two roots is p2, and from the second principle in the same article, this product is equal to q; hence, the supposition of p2 <q is absurd, and the values of the roots corresponding to the supposition ought to be im possible or imaginary. When any particular supposition gives rise to imaginary results, we interpret these results as indicating that the suppo sition is absurd or impossible. If p=0, the roots in each form become equal with con trary signs; real in the first and second forms, and imaginary in the third and fourth. If g 0, the first and third forms become the same, as also, the second and fourth. In the former case, the first root is equal to 0, and the second root is equal to 2p; in the latter case, the first root is equal to +2p, and the second to 0. If p = 0 and q = 0, all the roots in the four forms reduce to 0. In the preceding discussion we have made p2 > q, p2 <q, and p2 = 9; we have also made p and q separately equal to 0, and then both equal to 0 at the same time. These suppositions embrace every possible hypothesis that can be made upon p and 9. 118. The results deduced in article 117 might have been obtained by a discussion of the four forms themselves, instead of their roots, making use of the principles demonstrated in arti cle 116. In the first form the product of the two roots is equal to q, hence the roots must have contrary signs; their sum is -2p, hence the negative root is numerically the greater. Ч In the second form the product of the roots is equal to and their sum equal to +2p; hence, their signs are unlike, and the positive root is the greater. In the third form the product of the roots is equal to +q; hence, their signs are alike, and their sum being equal to -2p, they are both negative. In the fourth form the product of the roots is equal to +%, and their sum is equal to +2p; hence, their signs are alike and both positive. If p = 0, the sum of the roots must be equal to 0; or the roots must be equal with contrary signs. If q 0, the product of the roots is equal to 0; hence, one of the roots must be 0, and the other will be equal to the coefficient of the first power of the unknown quantity, taken with a contrary sign. If P = 0 and q 0, the sum of the roots must be equal to 0, and their product must be equal to 0; hence, the roots themselves must both be 0. 119. There is a singular case, sometimes met with in the discussion of problems, giving rise to equations of the second degree, which needs explanation. To discuss it, take the equation ax2 + bx = c, If, now, we suppose a = = 0, the expression for the value of ≈ becomes == But the supposition a = 0, reduces the given equation to bx = c, which is an equation of the first degree. The roots, found above, however, admit of interpretation. 0 0 The first one reduces to the form in consequence of the existence of a factor, in both numerator and denominator, which factor becomes 0 for the particular supposition. To deduce the true value of the root, in this case, take and multiply both terms of the fraction by -b-b2+4ac; after striking out the common factor x= 2c 2a we shall have b+ in which, if we make a = 0, the value of x reduces to the same value that we should obtain by solving the simple equation bec. The other root ∞, is the value towards which the expression, for the second value of x, continually approaches as a is made smaller and smaller. It indicates that the equation, under the supposition, admits of but one root in finite terms. This should be the case, since the equation then becomes of the first degree. 120. The discussion of the following problem presents most of the circumstances usually met with in problems giving rise to equations of the second degree. In the solution of this problem, we employ the following principle of optics, viz. :— The intensity of a light at any given distance, is equal to its intensity at the distance 1, divided by the square of that distance. 121. Find upon the line which joins two lights, A and B, of different intensities, the point which is equally illuminated by the lights. Let A be assumed as the origin of distances, and regard all distances measured from A to the right as positive. Let c represent the distance AB, between the two lights ; a the intensity of the light A at the distance 1, and b. the intensity of the light B at the distance 1. Denote the distance AC, from A to the point of equal illumination, by x; then will the distance from B to the same point be denoted by c-x. From the principle assumed in the last article, the intensity of the light A, at the distance 1, being a, its intensity at the In like manner, the intensity of B at the distance c b (c x, is but, by the conditions of the problem, these two x) 2 intensities are equal to each other, and therefore we have the equation Since both of these values of x are always real, we conclude that there will be two points of equal illumination on the line AB, or on the line produced. Indeed, it is plain that there should be, not only a point of equal illumination between the lights, but also one on the prolongation of the line joining the lights and on the side of the lesser one. To discuss these two values of x. First, suppose a >b. The first value of x is positive; and since it will be less than c, and consequently, the first point C, will be situated between the points A and B. We see, moreover, that the point will be nearer B than A; for, since a >ö, we have α √a + √ a or, 2√/a>(√a+b), whence |