REMARKS.-1. If any figure, found by trial, is too great or too small, it will be made manifest in the next transformation. (See Art. 431.)

2. After finding three figures of the root, the next three may generally be obtained by dividing the absolute term by the coefficient of x.

Find at least one value of x in each of the following:

15. x3+2x2+3x3+4x2+5x=54321. A. x=8.414455.

435. To extract the roots of numbers by Horner's Method.

This is only a particular case of the solution of the equation "N, or "-N=0; an equation of the nth degree, in which all the terms are wanting except the first and last.

In performing the operation, observe that the successive integral figures have the same relation to each other that the successive decimal places have in the previous examples.

In extracting any root, point off the given number into periods, as in the operation by the common rule.

For an example, let it be required to find the cube
root of 12977875; that is, one root of the equation
2-12977875=0.

The reason for placing the figures as they are in the successive
columns, will be readily understood by using the numbers 200 and
30, instead of 2 and 3.

5. The fifth root of 68641485507. . Ans. 147.

APPROXIMATION BY DOUBLE POSITION.

436. Double Position furnishes one of the most use-
ful methods of approximating to the roots of equations.
It has the advantage of being applicable, whether the equa-
tion is fractional, radical, or exponential, or to any other
form of function.

Let X=0, represent any equation; and suppose that a and b, substituted for x, give results, the one too small, and the other too great, so that one root lies between a and b. (Art. 403.)

Let A and B be the results arising from the substitution of a and b for x, in the equation X=0. Let x=a+h, and b=a+k; then, if we substitute a+h and a+k for x, in the equation X=0, we shall have

X=A+A'h+A"h2+, etc.
B=A+A'k+¿A′′k2+, etc.

Here, A', A", etc., are the derived functions of A (Art. 411). Now, if h and k be so small that their second and higher powers may be neglected without much error, we shall have

X-A-A'h nearly;
B-A=A'k 46

Whence, B-A : X—A : : A ́k : A'h: k: h;

B-A: k

:: X—A : h, (Art. 270);

B-A: b —a :: X—A : h, since k—b—a.

Hence, we have the following

Rule.-Find, by trial, two numbers which, substituted for x, give one a result too small, and the other too great. Then say, As the difference of the results is to the difference of the suppositions, so is the difference between the true and the first result, to the correction to be added to the first supposition.

Substitute this approximate value for the unknown quantity, and find whether it is too small or too great; then, take two less numbers, such that the true root may lie between them, and proceed as before, and so on.

It is generally best to begin with two integers which differ from each other by unity, and to carry the first approximation only to one place of decimals. In the next operation make the difference of the suppositions 0.1, and carry the 2d quotient to two places, and so on.

1. Given x3+x2+x=100, to find x.

Here, lies between 4 and 5. Substitute these two numbers

for x in the given equation, and the result is as follows:

therefore, x=4.2, the first approximation.

Substituting 4.2 and 4.3 for x, and proceeding as before, we get

for a second approximation_x=4.264.

Assuming x=4.264 and

4.265, and continuing, we obtain x=4.2644299, nearly.

Find one root of each of the following equations:

7. †/7x3+4x2+1/10x(2x−1)=28. A. x=4.51066

437. Newton's Method of Approximation.-This method, now but little used, is briefly as follows:

Find, by trial, a quantity a within less than 0.1 of the value of the root. Substitute a+y for x in the given equation, and it will be of this form

A+A'y+A”y2+}A""'y3+, etc., =0 (Art. 411),

where A, A', A", etc., are what the proposed equation, the first derived polynomial, etc., become when xa.

y is <0.1, y2 will be <0.01, y3<0.001, and so on.

Therefore, if the sum of the terms containing y2, y3, etc., be less than .01, we shall, in neglecting them, obtain a value of y within

Α
A

.01 of the truth. Putting y=- we have xa—·

differ from the true value of x by less than .01.

A

This will

A

Now, put b for this approximate value of x, and let x=b+z; we have then as before

B+B'z+B'z2+B23+, etc., =0;

and as z is supposed to be less than .01, 22 will be <.0001. If, then, we neglect the terms containing 22, 23, etc., we shall obtain a probable value of z within .0001; and so on. Applying the successive corrections, we obtain the value of x.

Newton gave but a single example, viz.:

Required to find the value of x in the equation x3-2x -5-0.

Ans. x

2.09455149.

CARDAN'S RULE FOR SOLVING CUBIC EQUATIONS.

438. In its most general form, a cubic equation may be represented by

x3+px2+qx+r=0;

but as we can always take away the second term, (Art. 407,) we will suppose, to avoid fractions, that it is reduced to the form

x3+3qx+2r=0.

Assume x-y+z, and the equation becomes

y3+z3+3yz(y+z)+3q(y+2)+2r=0.

Now, since we have two unknown quantities in this equation, and have made only one supposition respecting them, we are at liberty