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of the third form, whose roots become impossible.

2. It is obvious that, in the two first forms, one of the roots must be positive, and the other negative. 3. In the third form, if an, or the square

4 of half of the coefficient of the unknown quantity be greater than 63, the known quantity, the two roots will be positive. If a be equal to 62, the two roots then be

4 come equal. But if, in this third case, is less than

4 b>, the quantity under the radical sign becomes negative, and the two roots are therefore impossible. This. may be easily shewn to arise from an impossible supposition in the original equation.

4. If the equation, however, express the relation of magnitudes abstractly confidered, where a contrariety cannot be supposed to take place; the negative roots cannot be of use, or rather there are no such roots; for then a negative quantity by itself is un



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intelligible, and therefore the square root of a positive quantity must be positive only. Hence, in the two first cases, there will be only one root; but, in the third, there will be two.

For, in this third case, x?-ax= –62, or ax-x=b, it is obvious that « may be either greater or less than şa, and yet amxmay be positive ; and hence axXx=axaca


also be pofitive, and may be equal to a given positive quantity b2; therefore the square root of x-ax+ ta? may be either x-la, or {a-x, and both these quantities also positive. Let then x

-b?, and x= +

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are the same two positive roots as were obtained by the general rule.

The general rule is usually employed, even in questions where negative numbers cannot take place, and then the negative roots of the two first forms are neglected.


Sometimes even, only one of the positive roots of the third case can be used, and the other may be excluded by

be excluded by a particular condition in the question. When an impofsible root arises in the solution of a queftion, which has been resolved in general terms, the necessary limitation of the data will be difcovered.

When a question can be so stated, as to produce a pure equation, it is generally to be preferred to an adfected. Thus the question in the preceding section, by the most obvious notation, would produce an adfected equation.

II. Solution of Questions producing Quadra

tic Equations. The expression of the conditions of the question by equations, or the stating of it, and the reduction likewise of these equations, till we arrive at a quadratic equation, involving only one unknown quantity and its square, are effected by the same rules which were given for the solution of simple equations, in Chap. III.


One lays out à certain sum of money in goods,

which he fold again for L. 24, and gain-
ed as much per cent. as the goods cost him:
I demand what they cost him ?

per cent.

If the money laid out be tly
The gain will be

224 But this gain is


3 (y:24-y::100)


2400-rooy • Therefore by question,

у And by mult. and tr. 5!y: +10oy=2400 Completing the square 6 y? +100y+50=2400+2500=4900. Extr. the root

7y +50=+V4900=70 Trans,

8y=+70-50=20, or — 120.

4 y =

The answer is 20l. which succeeds. The other root, — 120, has no place in this example, a negative number being here unintelligible.

Any quadratic equation may be resolved also by the general canons at the beginning of this section. That arising from this question, (No. 5.), belongs to Cafe 1. and a=100, bạ=2400; therefore,

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+2400=20, or -120, as before.



What two numbers are those, whose diffe

rence is 15, and balf of whose product is equal to the cube of the lefser ?

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** +15%=x3


By question
Divide by * and mult. by 2. 4*+15=2x?
4th prepared


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The numbers, therefore, are 3 and 18, which answer the conditions. This is an example of Case 2d, and the negative root is neglected.


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