12. If equation Р

[(s2 —p3) +§]}

3-3px=-28 we shall find by reducing the

x==+u=

น

p

[{s2 — p3) }; — s]{}

13. If x+3px=28

p

By assuming x ==-u, we shall find

We see that this belongs to the first form of cubics.
Comparing it with that form, we find

Substituting these values of p and s for p and s, in the value of x found above, we have

15. Solve by the general formulæ the following equations:

By this rule we find one of the values of x; the method of finding the other two values is explained in Section LIII.

Before applying the formule to the two last examples their forms must be changed so that they may contain only the third and first powers of the unknown number. See the last section.

The means of effecting this change has been applied to examples (1,) (2,) (3,) (4,) at the end of the section referred to.

3

In finding the value of x in the first and second forms of cubics given above, we have to subtract p3 from s2, and extract the square root of the remainder. Now it may happen, that p3 is greater than s2, and then the result of this subtraction is a negative quantity or number; and, since the square root of a negative number cannot be found, we cannot, in such a case, find the value of x by the preceding rule. This is called the irreducible case of cubics. The reason of the failure of the rule may be easily seen. By the rule we find two parts of x, the product of which is equal to p, a given number; but, x may have such a value, that the product of no two of its parts can be equal to p; and, consequently, the two parts given by the rule, will be imaginary, although x, the sum of the two parts, is a real number.

No direct rule has been found for reducing a cube, when p is greater than s2. In such cases we have recourse to approximating rules, by which we can obtain the value of x to any required degree of exactness.

SECTION XLVIII.

On Cubic Equations.

If (x — a) (x —b) (x — c) = 0, a being greater than b, and b greater than c.

First. If we substitute for x, a number greater than a, the result will be positive. Why?

Second. If we substitute for x, a number between a and b, the result will be negative. Why?

Third. If we substitute for x, a number less than c, the result will be negative. Why?

Fourth. If we substitute two consecutive numbers, successively for x, and one result is +, whilst the other is; then we know that a true value of x is between the two numbers; and, consequently, that the least whole number substituted, is the whole number in a true value of x.

It may be easily shown, that this last proposition is true for equations of any degree.

SECTION XLIX.

Of the Resolution of Equations by Approximation.

The process in the following example may be applied to an equation of any degree.

1. Find the value of x in the following equation:

x3 + x2+x=90

We can see from a few trials, that the value of x is between 4 and 5; for, the third, second and first powers of 4 added together, give 84. The third, second and first powers of 5 added together, give 155; so that the exact value of x is between 4 and 5.

Assume
Then

4+2=x (4 + z)3 + (4+2)2 + 4 + z = 90 64+48z+12 z2 + z3 + 16 +8z+z2+4+z = 90 z3 + 13 z2 + 57 % + 84 = 90 2+13 z2+57 z = 6

Since z is a fraction, its third and second powers are very small, and may be rejected in the preceding equation; and 57 z = 6, nearly.

z=1', or .1, nearly. x=4+2=4.1, nearly.

Hence

Assume again,

≈=4.1+z

Substitute as before, and we shall find

z=.002

Hence x 4.1+z = 4.102, nearly

We might assume again,

x=4.102+z

Substitute again, and find a number still nearer the value of x.

We can always reject every power of z, except the first; since these powers are very small.

Find the values of x in the following equations:

SECTION L.

Second Method of Approximation.

We will give another method of approximation, which is more easily applied to some equations, than the preceding. 1. Find the value of x in the following equation: x3 + 4x2+6x= 100

Find by trial, two numbers near the value of x.

Substituting 3 for x, the first member of the equation becomes 81. If we had substituted the correct value of x, the first member would have been 100. Taking 81 from 100, and the remainder is 19, this we call the first error.

Substituting 3.1 for x, the first member becomes 86.831; consequently, 13.179 is the second error.

The difference of the numbers assumed for x is, the difference of the errors is 5.821. Now, this difference of the errors is owing to the difference in the assumed numbers.

If 5.821 give a difference of, what will 13.179 give? It is assumed that it will give a proportional number which is not strictly true; but, near enough to the truth for the purpose of approximation.

is 3.327 of 5.821. Hence, the difference in the errors, 5.821, gives a difference in the values assumed for x, equal to 3.21 of itself; 13.179 must give a difference of the values of 2 equal to T of itself; but of 13.179 is .22, nearly. Now 13.821 is produced by the difference between 3.1 and the true value of x. Hence, 3.32 is the true value of x,

nearly.

.1

Take 3.1 and 3.32, and substitute them separately for x; and note the results and the errors; and repeating the process, we shall find another number still nearer the value of x.

By repeating the process, we may find the value of x to any required degree of exactness.

When one of the numbers assumed for x, gives a number too great, and the other a number too small, we then use the sum of the errors instead of their difference, which we used above, the reason of which may be easily seen.

Find the values of x in the following:

SECTION LI.

Discussion of Cubic Equations.

Given, (x-a) (x —b) (x — c) = 0 to find the values of x. Any one of these factors may be = 0, hence we have

This equation gives three values of x, a, b and c.
The values of x are called the roots of the equation.
If the three factors be multiplied together, we have

x3-(a+b+c) x2+(ab+ae+bc) x—abc=0 This cubic equation is the same as equation (1,) only in a different form; and, if reduced, would give the same values of x. Every cubic equation may be conceived to be formed in the same way, when it is reduced to the preceding form, as every cubic equation may be. In every cubic equation, then, in which the first term is the third power of the unknown quantity,

1st. The second term is the second power of the unknown quantity, with a coefficient equal to the sum of the roots, with the contrary sign; for, the values of x, in the preceding equation, are all positive.

2nd. The third term is the first power of the unknown quantity, with a coefficient equal to the sum of all the products which can be made, by taking the roots two by two.* The three letters, a, b and c, may be combined three ways, ab, ac, bc.

3rd. The fourth term is the continued product of all the roots, with the contrary sign.

It is easy then, to form a cubic equation which, shall have any three given positive numbers for its roots.

2. Assuming x for the unknown quantity; what will the equation be which shall have 1, 2 and 3 for its roots?

Answer. (1 + 2 + 3) x2 + (2+3+6) x-6=0, or 23 6 x2+11x= 6.

3. What equations will have the following roots?

* It is evident that all the different products which can be made by taking the letters a, b, c, two by two, are a b, a c and b c.