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These principles are employed in discussing the general equation of the second degree.



147. We have seen that the discussion of an equation consists in making every possible supposition on the arbitrary quantities which enter it, and interpreting the results. In discussing the equation,

202 + 2pac = 9, we shall first make every possible hypothesis with respect to the signs of p and q. It is plain, first, that both p and q may be positive; second, that p may be nega. tive, and a positive; third, that p may be positive, and a negative; and fourth, that P

and negative. These hypotheses give,

9 may be

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2c2 + 2pc

(1.) 2? 2pac = 2

(2.) 2? + 2pac -2

(3.) 22 2poc = - 9

(4.) First Form. Since 2 is positive, the product of the roots must be negative (Principle 4); hence, the roots have contrary signs ; again, since 2p is positive, the algebraic sum of the roots is negative (Principle 3); hence, the negative root is numerically the greater.

Second Form. For the reason as before, the roots have con trary signs; since 2p is negative, the algebraic sum


of the roots is positive (Principle 3); hence, the positive root is numericully the greater.

Third Form

Here, is negative, and consequently the product of the roots is positive (Principle 4); hence, the roots have the same sign; and, since 2p is positive, the sum of the roots must be negative (Principle 3); hence, both roots are negative.

Fourth Form.



For the same

as before, the roots have the same sign; because 2p is negative, the sum of the roots is positive (Principle 3); hence, the roots both positive.

If now, we suppose p = 0, and a not 0, the first and second forms reduce to

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In this case, the roots have contrary signs, and their sum is equal to 0. Hence, they are both equal, with contrary signs.

Under the same supposition, the third and fourth forins become,

- 2

202 =

In this case, the roots have the same sign, and their sum is (, which is impossible. This case comes under the 5th principle.

Solving the equation, we find that both of the roots are imaginary. It will always be found, when the second member is negative, and numerically greater than the square of half the coefficient of the second term, that the roots are imaginary, as may be shown by solving either the third or fourth form, and making p2 < 2,

Imaginary roots always indicate some absurdity in the conditions that have given rise to the equation whence they are derived, and are to be so interpreted.

If we suppose q = 0, and p not equal to 0, the first and third forms reduce to

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In both cases, the product of the roots is equal to 0, which shows that one of them must be 0. In the first case, the second root is equal to 2p, and in the second, it is equal to + 2p.

If we suppose, p = 0, and q =- 0, all of the forms reduce to

22 = 0.

In this case, the product of the roots is 0, hence, one of them must be 0, and their sum is equal to 0; hence, the second root is also equal to 0.

The results of the above discussion might be arrived at, by solving the four forms, and discussing the roots thus obtained. Were we to solve them, we should arrive at no results not indicated above.


148. The problem of the lights affords a fair example of problems of the second degree, and its discussion indicates the general method of reasoning in such cases.

It is a principle of Physics, that the intensity of a light, at any distance, is equal to its intensity at the distance 1, divided by the square of that distance.


Having given the intensities of two lights, at the dis tance 1, and the distance between the lights, it is required to find that point on the line joining them which is equally illuminated by them.

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Let A and B represent the places of the two lights, and AB the line joining them. Assume A, as the origin of distances, and call all distances estimated to the right positive, then will all distances to the left be negative.

Denote the distance between the lights by C, the intensity of the light A, at the distance 1, by a, and that of the light B, at the distance 1, by b. Suppose the point R, to be equally illuminated, and denote its distance from the origin A, by x; then will the distance BR, be equal to c

Since the intensity of the light A, at the distance 1, is a, at the distance x, it will be in accordance with the physical principle enunciated; and since the intensity of the light B, at the distance 1, is b, at the distance

ń . – X, it will be equal to

From the conditions

(c - 2)2 of the problem, these two expressions must be equal; hence,


(c − x) 6


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(c — x)2; or,

Extracting the square root of both members, we have,

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From the nature of the problem, both a and b are positive, and the two values of x are both real.

Hence, there are two points of equal illumination, and only two, on the line of the lights.

To discuss these values of x, we observe that c, d, and b, are, from the nature of the case, arbitrary. They are also positive; and since the conditions of the problem in. volve the necessity of two lights, neither a, b, nor c, can be 0. Hence, only three different suppositions, belonging to the problem, can be made on these arbitrary quantities, viz. :

1. c>0, and a > b. 3. c> 0, and a < b. 2. c> 0, and

a = b.

1. c>0, and a > b.

is a proper

va In this case, the fraction,

Va + vb' fraction, and therefore less than 1; and because the denominator is less than twice the numerator, it is greater than }; hence, the first value of x is less than c, and greater than {c: which shows that the first point is between the two lights, and nearer the lesser one. The fraction,


is improper, and therefore -

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