Here q+ 4 will evidently be positive; the value of x may therefore be obtained, either exactly or with such degree of approximation as we please. With respect to the two values of x, the first, viz. p2 will be positive, for the square root of alone being the 4 2' square root of q+will be greater than the value of z will therefore have the same sign with the radical and will by consequence be positive. This value will answer directly the conditions of the equation, or the problem of which the equation is the algebraic translation. The second value of x, viz. x= 干 P 2 p2 9+ being 4' also necessarily of the same sign with the radical, will be essentially negative. This value, though it satisfies the equation, will not answer the conditions of the question, from which the equation is derived. It belongs to an analogous question corresponding to the equation, after troduced instead of x, that is, to x3 px = q. p this last equation, we deduce x=±1⁄2 ± | has been inIndeed, from which do not differ from the preceding except in the sign. Thus the same equation connects together two questions, which differ from each other only in the sense of certain conditions. 2. Again, let q be negative. The equation will then be of the form xpx=- - q, and we have Here in order that the root of the quantity placed under the radical sign may be taken, or in other words, that the value of I may be real, it is evident that q must not exceed p2 p 2' Since moreover q is numerically less than it follows, that the values of x will both be negative, if p is positive in the equation, that is, if the equation is of the form x2 + p x = — q; and that they will both be positive, if p is negative in the equation, that is, if the equation is of the form r2 ра x=- q. Indeed, it may be shown a priori, that always when q is negative, in the second member and p negative in the first, the problem will admit of two direct solutions, provided that ૧ does ,p2 not exceed 4. The equation x2-px-q may, by changing the signs of all the terms, be put under the form But the equation x (px)=q is evidently the algebraic translation of the following enunciation, viz. To divide a number p into two parts, the product of which shail be equal to a given number q. For if we put x for one of the parts, the other part will be p—x, and the product of the two parts, will be x (px). This being premised, the enunciation of the problem admits, it is evident, of two direct solutions; for the equation of the problem will be the same, whether x be put for one or the other of the parts; there is no reason then, why the equation, when resolved, should give one of the parts rather than the other; it should therefore give both at the same time. Moreover, in order that the problem may be possible, it 2 p2 is necessary, that q should not exceed 4 for the greatest ; possible product of the parts, into which the number p may be p2 divided being equal only to it is absurd to require that their product, which we have represented by q, should be greater than p2 4 We conclude therefore that, in all cases when the known quantity is negative in the second member, but numerically greater than the square of half the coefficient of the second term, the question proposed is impossible. 120. The following examples will serve as an exercise upon the different cases, which we have here been considering. What change must be made in the enunciations of the first five questions respectively, in order that the negative solutions may become positive? How must the sixth question be modified, so that the answers shall become positive? In what does the absurdity in the ninth question consist? 1. A company at a tavern had £8, 15s. to pay, but two of them having left before the bill was settled, those who remained had each in consequence 10s. more to pay. How many were in the company at first? 2. A man travelled 105 miles, and then found that if he had not travelled so fast by 2 miles an hour, he should have been 6 hours longer in performing the same journey. How many miles did he go per hour? 3. A regiment of foot was ordered to send 216 men on garrison duty, each company being to furnish an equal number; but before the detachment marched, 3 of the companies were sent on another service, when it was found that each company that remained was obliged to furnish 12 additional men, in order to make up the complement 216. How many companies were there in the regiment, and what number of men was each ordered to send at first? 4. A and B set out from two towns, which were distant 247 miles, and travelled the direct road till they met. A went 9 miles a day; and the number of days at the end of which they met, was greater by 3, than the number of miles, which B went in a day. Where between A and B did they meet? 5. A merchant sold a commodity for 56 dollars, and gained as much per cent. as the whole cost him. What was the cost? On substituting-x for x in the equations, which pertain respectively to the preceding questions, it will be easy to translate these equations into enunciations analogous to those of the questions proposed; there are questions however, in which it will be very difficult to do this, and the negative, solutions in such cases are to be regarded merely as connected with the first in the same equation of the second degree. 6. A gentleman counting the guineas, which he had in his purse, finds that if 24 be added to their square, and 8 times their number be subtracted from 17, the sum and remainder will be equal. How many guineas had he in his purse? 7. A set out from C towards D, and travelled 7 miles a day. After he had gone 32 miles, B set out from D towards C, and went every day one-nineteenth of the whole journey; and after he had travelled as many days as he went miles in one day, he met A. Required the distance of the places C and D ? 8. Two merchants sell each a certain number of yards from the same piece of cloth, the second three yards more than the first, and received jointly $35. Said the first to the second, at my price I should have received $24 for your cloth, and I, replies the other, at my price should have received only $124 for your cloth. How many yards did each sell? 9. The difference of two numbers is 7, and the square of the greater is equal to 25 times the less; What are the numbers? EXAMINATION OF PARTICULAR CASES. 1. In the general equation let q be negative, that is, let the equation be of the form x2 + px 9, p being of any sign whatever; if we suppose q the radical will be reduced to 0, and the values of x will be equal each to -Thus if q be negative in the equation and equal to 2 the values of x will be equal, and will both be positive if negative, or both negative if p is positive. 2. In the general formula let q 3. In the same formula let p 0, we have then x= 4' is that is to say, the values of x will in this case be equal, but of contrary signs, real if q is positive, and imaginary if q is negative. 4. Let p=0, q=0, the values of x will then be each equal to 0. 5. We have next to examine a remarkable case which frequently occurs in the solution of problems of the second degree. For this purpose, let us take the equation Let it now be supposed, that in consequence of a particular hypothesis made upon the given things in the question, we have a 0, the values of x then become The second value of x here presents itself under the form of infinity and may be regarded as a true answer, when the question is susceptible of infinite solutions. In order to interpret the first, if we return to the equation, we see that the hypothesis a―0 reduces it to b x c, from which we deduce be regarded as the true value of in the present case. 0 6. Let it be supposed finally, that we have at the same time a = 0, b = 0, c=0. The equation will then be altogether indeterminate. This is the only case of indetermination, which the equation of the second degree presents. DISCUSSION OF PROBLEMS. 121. The following problems offer all the circumstances, which usually occur in problems of the second degree. 1. To find on the line A B, which joins two luminous bodies A and B, the point where these bodies shine with equal light. The solution of this problem depends upon the following principle in physics, viz. The intensity of light from the same |