The Project Gutenberg eBook #36640: Lectures on Elementary Mathematics

M ₊ N _.

x² (a − x)²

Directions for finding the smallest and greatest values of variable quantities are given in the Differential Calculus. We shall here content ourselves with remarking that the quantity in question will be a minimum when

x _{= 3} M _;

a − x N
.

so that we shall have

a^√3 M

x = ^√3 _{M +} ^√3 _N ,
from which we get, as the smallest value of the expression
M ₊ N _,

x² (a − x)²

the quantity (_√3
M +


^√3 N )³

_a2 .

Hence there will be two real values for x if this quantity is less than A; but these values will be imaginary if it is greater. The case of equality will give two equal values for x.

I have dwelt at considerable length on the analysis of this problem, (though in itself it is of slight importance,) for the reason that it can be made to serve as a type for all analogous cases.

The equation of the foregoing problem, having been freed from fractions, will assume the following form:

Ax²(a − x)² − M (a − x)² − Nx² = 0.

With its terms developed and properly arranged it will be found to be of the fourth degree, and will consequently have four roots. Now by the analysis which we have just given, we can recognise at once the character of these roots. And since a method may spring from this consideration applicable to all equations of the fourth degree, we shall make a few brief remarks upon it in passing. Let the general equation be

x⁴ + px² + qx + r = 0.

We have already seen that if the last term of this equation be negative it will necessarily have two real roots, one positive and one negative; but that if the last term be positive we can in general infer nothing as to the character of its roots. If we give to this equation the following form

(x² − a²)² + b(x + a)² + c(x − a)² = 0,

a form which developed becomes

x⁴ + (b + c − 2a²)x² + 2a(b − c)x + a⁴ + a²(b + c) = 0,

and from this by comparison derive the following equations of condition

b + c − 2a² = p, 2a(b − c) = q, a⁴ + a²(b + c) = r,
and from these, again, the following,

b + c = p + 2a², b − c = ^q , 3a⁴ + pa² = r,
2a

we shall obtain, by resolving the last equation,

₂ p ^. r p²
= − ₆ +

3 ⁺ 36 ^.