I shall show that, if there exists a quantity a, which, substituted in the place of x, verifies the equation of the second degree, x2 + px = q, and is consequently the value of x, this unknown quantity will still have another value. Now, if we substitute a for a, the result will be a2 + pa=q; and since, by supposition, a represents the value of x, q will be necessarily equal to the quantity a2 + pa; we may then write this quantity in the place of q, in the proposed equation, which thus becomes x2 + px = a2 + pa. Transposing all the terms of the second member, we have it is obvious, at once, that the first member is divisible by r — -a, and will give an exact quotient, namely, x + a +p; we have then, x2 + px-q= x2 — a2 + p (x — a) = (x — a) (x+a+p). Now it is evident, that a product is equal to zero, when any one of its factors whatever becomes nothing; we shall have then but also when x + a + p = 0, from which is deduced This result agrees with the two values comprehended in the formula for if we take for a the first value, - ≤ p + √q + { p2, we obtain for the other α P —a—p = + P — √ q + { p2 —p➖➖1p-√ q + ÷ p2, = +‡p2, which is in fact the second value. These remarks contain the germ of the general theory of equations of whatever degree, as will appear hereafter, when the subject will be resumed. 117. The difficulty of putting a problem into an equation, is the same in questions involving the second and higher powers, as in those involving only the first, and consists always in disentangling and expressing distinctly in algebraic characters all the conditions comprehended in the enunciation. The preceding questions present no difficulty of this sort; and, although the learner is supposed to be well exercised in those of the first degree, I shall proceed to resolve a few questions, which will furnish occasion for some instructive remarks. A person employed two labourers, allowing them different wages; the first received, at the end of a certain number of days, 96 francs, and the second, having worked six days less, received only 54 francs; if this last had worked the whole number of days, and the other had lost six days, they would both have received the same sum; it is required to find how many days each worked, and what sum each received for a day's work. This problem, which at first view appears to contain several unknown quantities, may be easily solved by means of one, because the others may be readily expressed by this. If a represent the number of days' work of the first labourer, 6 will be the number of days' work of the second, X 96fr. x 54 will be the daily wages of the first, X- 6 the daily wages of the second; if this last had worked x days, he would have earned The first step is to make the denominators disappear; the equation then becomes As the numbers 54 and 96 are both divisible by 6, the result may be simplified by division, we shall then have 9 x 2 = 16 (x 6) (x — 6). This last equation may be prepared for solution according to the rule given art. 103, but as the object of this rule is to enable us with more facility to extract the root of each member of the equation proposed, it is here unnecessary, because the two members are already presented under the form of squares; for it is evident, that 9 x2 is the square of 3 x, and 16 (x-6) (x — 6) of 4 (x-6). We have then the square By the first solution, the first labourer worked 24 days, and consequently earned or 4 francs per day, while the second worked only 18 days, and received 4 or 3 francs per day. The second solution answers to another numerical question, connected with the equation under consideration, in a manner analogous to what was noticed in art. 111. 118. A bunker receives two notes against the same person; the first of 550 francs, payable in seven months, the second of 720 francs, payable in four months, and gives for both the sum of 1200 francs ; it is required to find, what is the annual rate of interest, according to which these notes are discounted. In order to avoid fractions in expressing the interest for seven months and four months, we shall represent by 12 x the interest of 100 francs for one year; the interest for one month will then be x. The present value of the first note will accordingly be found by the proportion, 100+7x 100 :: 550 : 55000 100+7x (Arith. 120); and the present value of the second note by the proportion, By uniting these values, we obtain for the equation of the problem, 55000 + 72000 = 1200. 100+7x 100+ 4x Dividing each of the members by 200, we have making the denominators disappear, we find successively, 275 (100 + 4 x) + 360 (100 + 7 x) = 6 (100 + 7 x) (100 + 4 x), 27500 + 1100x + 36000 + 2520x = 60000 + 6600x + 168x2, which may be reduced to If we then, since the denominator of this fraction is a perfect square, we have only to extract the square root of its numerator. stop at thousandths, we find 837,869, for the root of 702025; this, taken with the denominator 84, gives for the values of x The first of these values is the only one, which solves the question in the sense, in which it was enunciated. Dividing the denominator of this fraction by 12, we have (Arith. 54.) that is, the annual interest is at the rate of 13,27 nearly. 119. The following question deserves attention on account of the character, which the expression for the unknown quantity presents. To divide a number into two parts, the squares of which shall be in a given ratio. Let a be the given number, m the ratio of the squares of its two parts, x one of these parts; We shall then have, according to the enunciation, This may be resolved in two ways; we may either reduce it to the form x2 + px = q, and then resolve it by the common method; or since the fraction. is a square, the numerator and denominator being each a square, we thence conclude at once, x ax x = ± (α − x) √/m• By resolving separately the two equations of the first degree comprehended in this formula, namely, By the first solution, the second part of the number proposed is 1 + √ m 1 + √ m are both, as the enunciation requires, less than the number pro |