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second degree, where q is negative in the second member, the on

p2

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ly equations, which produce imaginary roots, since the term placed under the radical sign, preserves always the sign +, whatever may be that of p. Indeed, it is evident that the equation − q, or x2 + px + q = 0,

x2 + px = will admit of no positive solution, since the first member contains only affirmative terms; and, to ascertain whether the unknown quantity a can be negative, we have only to change x into y. The unknown quantity y would then have positive values, which would be furnished by the equation

y2 —py + q = 0, or y2-py=— q,

which is precisely the same as that in the preceding article; but as the values of x can be real, only when those of y would be so, they become therefore imaginary in the case under consideration, when q exceeds 22.

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It will be perceived then from what has been said, how, and for what reason, when the known term of an equation of the second degree is negative in the second member, and greater than the square of half the coefficient of the first power of the unknown quantity, this equation can have only imaginary roots.

115. The expressions

√=b, a + √ —b,

and, in general, those, which involve the square root of a negative quantity, are called imaginary quantities.* They are mere symbols of absurdity, that take the place of the value, which we should have obtained, if the question had been possible.

They are not, however, to be neglected in the calculation, because it sometimes happens, that when they are combined according to certain laws, the absurdity disappears, and the result becomes real. Examples of this kind will be found in the Supplement to this treatise.

116. As it is important, that learners should have just ideas respecting all those analytical facts, which appear to be derived from familiar notions, I have thought it proper to add some observations to what has been said (106), on the necessity of admitting two solutions in equations of the second degree.

* It would be more correct to say, imaginary expressions, or symbols, as they are not quantities.

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 x, 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 in the proposed equation, which thus becomes

x2 + px = a2 + pa.

Transposing all the terms of the second member, we have

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it is obvious, at once, that the first member is divisible by x-a, and will give an exact quotient, namely, x + a +p; we have then,

x2 + p x − 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 (x − a) (x + a + p) = 0,

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a = 0, which gives

x = α,

but also when x + a + p = 0, from which is deduced

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Therefore, if a is one of the values of x,-a-p will necessarily be the other.

This result agrees with the two values comprehended in the formula

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for if we take for a the first value, p + √q+÷p2, we obtain for the other

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These remarks contain the germ of the general theory of equa- ! tions 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 casily 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, x-6 will be the number of days' work of the second,

96fr.

54

will be the daily wages of the first,

x 6

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the daily wages of the second;

if this last had worked x days, he would have earned

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and the first, working x 6 days, would have received only

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The first step is to make the denominators disappear; the equation then becomes

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As the numbers 54 and 96 are both divisible by 6, the result may be simplified by division, we shall then have

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This last equation may be prepared for solution according to the rule given art. 108, 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) the square of 4 (x-6). We have then

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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 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 banker 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 z. The present value of the first note will accordingly be found by the proportion,

100+ 7 x 100 :: 550 :

55000 100+7x

(Arith. 120);

and the present value of the second note by the proportion,

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By uniting these values, we obtain for the equation of the problem,

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Dividing each of the members by 200, we have

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making the denominators disappear, we find successively,

275 (100+ 4x) + 360 (100+7x) = 6 (100 +7x) (100 + 4 x), 27500 + 1100x + 36000 + 2520x = 60000 + 6600x + 168x2, which may he reduced to

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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

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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.)

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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.

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