When the unknown quantity a represents the line AC", the

we have x>a; but by changing the signs in the equation

Therefore in this hypothesis, the point C, situated between A and B, must be nearer A than B.

gative. To interpret it, let us take for the unknown quantity the distance AC", and let us represent this distance by x, and at the same time consider, as we have a right to do, a as essentially negative. Then the general expression for BC" being a-x, if we regard x as essentially negative, the true numerical value of a-x is expressed by a+x. Hence as before, the equation or algebraic expression will be

or

x2 (a—x)2 x2 = (a+x)2

in the first of which equations x is essentially negative.

This equation ought to give a negative value for x, and a positive value for BC"=a+x. Indeed, since the intensity of the light B is greater than that of A, the second required point ought to be

The first two values of x and a-x reduce to

2'

which gives

the middle of AB for the first required point. This result agrees

with the hypothesis.

The two other values reduce to

±a√b
0

or infinity; that is, the second required point is situated at a distance from the two points A and B, greater than any assignable quantity. This result agrees perfectly with the present hypothesis, because, by supposing the difference b-c to be extremely small, without being absolutely nothing, the second point must be at a very great distance from the

lights; this is indicated by the expression

the denomi

nator of which is extremely small with respect to the numerator. And if we finally suppose b=c, or vb-c=0, the required point cannot exist for a finite distance, or is situated at an infinite distance. We will observe, that in the case of b=c, if we should consider the values before they were simplified, viz.

ence of a common factor, √b- ✔c, between the two terms of the value of x (see Art. 113).

Let b=c, and a=0.

The first system of values for x and a-x, reduces to 0, and the

0 second to This last symbol is that of indetermination; for, 0 resuming the equation of the problem, (b—c)x2—2abx=—a2b, it reduces, in the present hypothesis to 0.2-0.x=0, which may be satisfied by giving x any value whatever. In fact, since the two lights have the same intensity, and are placed at the same point, they ought to illuminate equally each point of the line A B.

The solution 0, given by the first system, is one of those solutions in infinite numbers, of which we have spoken.

Finally, suppose a=0, and b and c, unequal

Each of the two systems reduces to 0, which proves that there is but one point in this case equally illuminated, and that is the point in which the two lights are placed.

In this case, the equation reduces to (b-c)2=0, and gives the two equal values, x=0, x=0.

The preceding discussion presents another example of the precision with which algebra responds to all the circumstances of the enunciation of a problem.

Of Equations of the Second Degree, involving two or more unknown quantities.

151. A complete theory of this subject cannot be given here, because the resolution of two equations of the second degree involv. ing two unknown quantities, in general depends upon the solution of an equation of the fourth degree involving one unknown quantity; but we will propose some questions, which depend only upon the solution of an equation of the second degree involving one unknown quantity.

1. Find two numbers such that the sum of their products by the respective numbers a and b, may be equal to 2s, and that their product may be equal to p.

Let x and y be the required numbers, we have the equations,

This problem is susceptible of two direct solutions, because s is evidently-abp, but in order that they may be real, it is necessary that s2> or =abp.

Let a=b=1; the values of x, and y, reduce to

x=s±√s2-p and y=s√s2-p

Whence we see, that the two values of x are equal to those of y, taken in an inverse order; which shows, that if s+ √s2-p represents the value of x, s- √s-p will represent the corresponding value of y, and reciprocally.

This circumstance is accounted for, by observing, that in this par

ticular case the equations reduce to

(x+y=28,

xy=p;

and then the

question is reduced to, finding two numbers of which the sum is 2s, and their product p, or in other words, to divide a number 2s, into two such parts, that their product may be equal to a given number p.

2. Find four numbers in proportion, knowing the sum 2s of their extremes, the sum 2s' of the means, and the sum 4c2 of their squares.

Let u, x, y, z, denote the four terms of the proportion; the equa. tions of the problem will be

u+z=2s

x+y=28

uz=xy

u2+x2+ y2+z2=4c3.

At first sight, it may appear difficult to find the values of the unknown quantities, but with the aid of an unknown auxiliary they are easily determined.

Let P be the unknown product of the extremes or means, we have

Hence, we see that the determination of the four unknown quantities depends only upon that of the product p.

Now, by substituting these values of u, x, y, z in the last of the equations of the problem, it becomes.

2

(s+ √s2—p)2+(s— √s2—p)2+(s′+ √s'2—p)3

or, developing and reducing,

+(s'— √s'—p)2=4c2;

4s2+4s2-4p=4c; hence ps2+s'2-c2.

Substituting this value for p, in the expressions for u, x, y, z, we

These four numbers evidently form a proportion; for we have

u z=(s + √ c2 — s12) (s — √c2 —s2)=s2 —c2+s'2,
xy= (s' + √ c2 —s3) (s'′— √c2—s3)—s'2 — c2+s3.