Xn

305. As all the real roots of the proposed equation are necessarily included between and +∞, we may, by ascertaining the number of variations lost by the substitution of these, in succession, in the expressions X, X1 &c., readily determine the total number of such roots. It should be ob served, that it will be only necessary to make these substitutions in the first terms of each of the expressions, as in this case the sign of the term will determine that of the entire expression (Art. 282).

Having found the number of real roots, if we subtract this number from the highest exponent of the unknown quantity, the remainder will be the number of imaginary roots (Art. 248).

306. Having thus obtained the total number of real roots, we may ascertain their places by substituting for x, in succession, the values 0, 1, 2, 3, &c., until we find an entire number which gives the same number of variations as +∞. This will be the smallest superior limit of the positive roots in entire numbers.

Then substitute 1, 2, &c., until a negative number is obtained which gives the same number of variations as This will be, numerically, the least superior limit of the negative roots in entire numbers. Now, by commencing with this limit and observing the number of variations lost in passing from each number to the next in order, we shall discover how many roots are included between each two of the consecutive numbers used, and thus, of course, know the entire part of each root. The decimal part may then be sought by some of the known methods of approximation.

The first derived polynomial (Art. 264), is

24x2 6,

and since we may omit the positive factor 6, without affecting the sign, we may write

Dividing by X1, we obtain for the first remainder, -4x-1. Changing its sign, we have

4x + 1 = X2.

Multiplying X, by the positive number 4, and then dividing by X2, we obtain the second remainder

3; and by changing

its sign

+ 3 = X.

The expressions to be be used are then

X=8x3-6x-1, X1 = 4x2-1, X, 4x + 1, X = +3.

=

∞ and then, we obtain the two following

Substituting arrangements of signs:

there are then three real roots.

If, now, in the same expressions we substitute 0 and + 1, and then 0 and 1, for x, we shall obtain the three following arrangements:

As x1 gives the same number of variations as +∞,

1 gives the same as

and x = ∞, +1 and 1 are the smallest limits in entire numbers. In passing from

-1 to 0,

+1, one

two variations are lost, and in passing from 0 to variation is lost; hence, there are two negative roots between

If we deduce X, X1, and X, we have the three expressions

=

=

4x3 - 13x + 5,

= 13x2 15x + 38.

If we place X, 0, we shall find that both of the roots of the resulting equation are imaginary; hence, X, will be positive for all values of x (Art. 290). It is then useless to seek for X, and X.

3

hence, there are two real and two imaginary roots in the proposed equation.

307. In the preceding discussions we have supposed the equations to be given, and from the relations existing between the co-efficients of the different powers of the unknown quantity, have determined the number and places of the real roots; and, consequently, the number of imaginary roots.

In the equation of the second degree, we pointed out the relations which exist between the co-efficients of the different powers of the unknown quantity when the roots are real, and when they are imaginary (Art. 116).

Let us see if we can indicate corresponding relations among the co-efficients of an equation of the third degree.

Let us take the equation,

x3 + Px2 + Qx + U = 0,

and by causing the second term to disappear (Art. 263), it will take the form,

x3 + px + q = 0.

In order that all the roots be real, the substitution of ∞ for in the above expressions must give three permanences; and the substitution of - for x must give three variations. But the first supposition can only give three permanences when

- 4p3-27q2>0;

that is, a positive quantity, a condition which requires that p be negative.

or,

If, then, p be negative, we have, for x = ∞,

hence,

negative, and that

p3

27

conditions which indicate that the

roots are all real.

Cardan's Rule for Solving Cubic Equations.

308. First, free the equation of its second term, and we have the form,

or, by transposing, and substituting x for y+z, we have

and by comparing this with equation (1), we have

(2);

Solving this trinomial equation (Art. 124), we have

This is called Cardan's formula.

By examining the above formula, it will be seen, that it is inapplicable to the case, when the quantity

92 p3

+
4 27'

under the radical of the second degree, is negative; and hence, is applicable only to the case where two of the roots are imag inary (Art. 307).

Having found the real root, divide both members of the given equation by the unknown quantity, minus this root (Art. 247); the result will be an equation of the second degree, the roots of which may be readily found.