never be greater than

pa
4

The conditions of the equation there

pa fore fix a limit to the value of q, and if we make q> we express 4'

by the equation a condition which cannot be fulfilled, and, this contradiction is made apparent by the values of a becoming imaginary. Hence we may conclude that,

The value of the unknown quantity will always be imaginary when the conditions of the question are incompatible with each other.

REMARK. Since the roots of the equation, in the first and second forms, have contrary signs, the condition that their sum shall be equal to a given number p, does not fix a limit to their product : hence, in those two forms the roots are never imaginary.

148. We will conclude this discussion by the following remarks.

1st. If in the third and fourth forms, we suppose q=

p3
4

the ra

dical part of the two values of x becomes 0, and both of these

In this case, the first member is the product of two equal factors. Hence we may also say, that the roots of the equation are equal, since in this case the two factors being placed equal to zero, give the same value for x.

2d. If, in the general equation, x2+px=q, we suppose q=0,

In fact, the equation is then of the form x2+px=0, or x(x+p)=0, which can be satisfied either by supposing x=0, or x+p=0, whence x=-p: that is, one of the roots is 0, and the other the co-efficient of x taken with a contrary sign.

3d. If in the general equation x2+px=q, we suppose p=0, there will result x2=q, whence x=±√q; that is, in this case the two values of x are equal, and have contrary signs, real in the first and second forms, and imaginary in the third and fourth.

The equation then belongs to the class of equations involving two terms, treated of in (Art. 139).

4th. Suppose we have at the same time p=0, q=0; the equation reduces to x2=0, and gives two values of x, equal to 0.

149. There remains a singular case to be examined, which is often met with in the resolution of problems of the second degree. To discuss it, take the equation ax2+bx=c. This equation gives

Suppose now, that from a particular hypothesis made upon the given quantities of the question, we have a=0; the expression for x becomes

The second value is presented under the form of infinity, and may be considered as an answer when the proposed questions will admit of answers in infinite numbers.

0

As to the first

0'

we must endeavour to interpret it.

By multiplying the numerator and denominator of the 2d member of the equation

by

Hence we see that the indetermination arises from a common factor in the numerator and denominator.

If we had at the same time a=0, b=0, c=0, the proposed equation would be altogether indeterminate.

This is the only case of indetermination that the equation of the second degree presents.

We are now going to apply the principles of this general discussion to a problem which will give rise to most of the circumstances which are commonly met with in problems of the second degree.

150. Find upon the line which joins two lights, A and B, of different intensities, the point which is equally illuminated; admitting the following principle of physics, viz.: The intensity of the same light at two different distances, is in the inverse ratio of the squares of these distances.

Let the distance AB between the two lights be expressed by a; the intensity of the light A, at the units distance, by b; that of the light B, at the same distance, by c. Let C be the required point, and make AC=x, whence BC=a-x.

From the principle of physics, the intensity of A, at the unity of distance, being b, its intensity at the distances 2, 3, 4, &c., is b b b

&c., hence at the distance x it will be expressed by

с

In like manner, the intensity of B at the distance a-x, is

(a—2)a ;. but, by the enunciation, these two intensities are equal

to each other, therefore we have the equation

This expression may be simplified by observing, 1st. that b± √bc can be put under the form √b. √b±√b. √c, or √b(√b±√c); 2d. that b-c=( √b)3—( √c)2=(√b+ √c).( √b-√c.) Therefore, by first considering the superior sign of the above expression, we have

=

(√b+ √c).(√b— √c) vb-vc

In like manner we obtain for the second value,

The first value of x,

a, because

is a proper fraction; thus this value gives

for the required point, a point C, situated between the points A and B. We see moreover, that the point is nearer to B than A; for since bc, we have √b+ √b or 2 √b>(√b+√c); whence

ought to be the case, since the intensity of A was supposed to be

1. Hence this second value gives a

than a; because second point C', situated upon the prolongation of AB, and to the right of the two lights. We may in fact conceive that the two lights, exerting their influence in every direction, should have upon the prolongation of AB, another point equally illuminated; but this point must be nearest that light whose intensity is the least.

We can easily explain, why these two values are connected by the same equation. If, instead of taking AC for the unknown quantity x, we had taken AC', there would have resulted BC'=x-a;

and the equation

(x—a)3°

Now, as (x-a) is identical with

(a-x), the new equation is the same as that already established, which consequently should have given AC' as well as AC.

And since every equation is but the algebraic enunciation of a problem, it follows that, when the same equation enunciates several problems, it ought by its different roots to solve them all.