3. Given x3+9x2+4x=80, to find x.

Result, x=2.4721359.

4. Given x2+x2+x=90, required the value of x.

1

Ans. x 4.1028323.

5. Given x2+10x2+5x=2600, required the value of x.

6. Given value of x.

Ans. 11.00679934.

2x-16x3+40x2+30x=110, required the

N2

Ans. x=5.0569081.

127. To show the rationale of the process directed in the last article, I begin with the cubic equation

ax3+bx+cx+d=0.

Let the first figure in the root be indicated by r, regard being paid to its local value, and the remaining part

of the root by y; then x=y+r.

Hence, ax3=ay3+3ary3+3ar2y + ar3

bx=

by2+2bry+brz

=0.

Collecting the co-efficients of like powers of y,

ay3+by+c'y+d'=0.

It is obvious that d'={(ar+b)r+c}r+d;

c=(ar+b)r+(ar+b)r+ar+c; b=ar+ar+ar+b.

But these are the quantities found by the fourth precept.

Again, since r>y, cy+d' approximates to 0, or y= -d' nearly; but this is the mode prescribed in the fifth precept for finding the next figure of the root.

Denoting the number obtained by the last operation by s, and the remaining part of y by z, so that y=2+8, we shall obtain a new equation az3+b"z2+c"z+d"=0, in which d"={(as+b')s+c'}s+d'; c"=(as+b')s+(as+b') s+as+c'; b"=as+as+as; whence an approximate value of z, or a new figure of the root is manifestly deducible from this new equation, as before.

Assuming now the general equation

ax+bxn-1..... mx2+nx+p=0.

And denoting as before, the first figure of the root by

*, and making x=y+r, we have

Or ayn+b'yn--.. &c. n'y+p'=0, in which the quantities b', n', p', are composed of the co-efficients a, b, &c. and the powers of r combined, as directed in the fourth

precept. And here as before, y=-2, nearly.

n'

Hence it is obvious, that the successive figures of the root may be obtained by the same kind of process; whatever may be the index of the highest power of the unknown quantity.

INDETERMINATE PROBLEMS.

128. When a problem is given, in which the number of unknown quantities employed, is greater than the number of independent equations furnished by the conditions of the problem; one, at least, of those quantities, may be assumed at pleasure. Such problems, therefore, generally admit of an indefinite number of answers. There are, however, certain conditions sometimes annexed, by which the number of answers is partially limited. For example, the answers are required to be whole positive numbers, or they are required to be square or cube numbers. Such problems are termed indeterminate, or unlimited, though in some instances, each unknown quantity admits of but one value. If simple powers only of the unknown quantities are included in the equations, the problem is said to be of the first degree.

CASE 1.

(

To find the values of x and y, in whole positive numbers, from the equation ax=by+c; a, b, c, being given numbers, positive or negative.*

by+c

Here x= , and as x is to be a whole number,

by+c

a

a

must also be a whole number.

Now, if this quantity be multiplied by a whole number, the product must, evidently, be a whole number; also the sum or difference of this quantity, or either of its multiples, and any whole number, must necessarily be a whole number. Let, therefore, by + be thus changed

till we obtain y+c y+c=wh,t which

a

α

put =p;t hen y=ap-c',

a quantity that must be a whole number, because a, p, and c' are whole numbers. And this value being subby+c, the value of z will be obtained in terms of p, and given numbers.

stituted for y in the equation x=

Assuming then p=0, 1, 2, 3, &c. successively, (omiting such numbers as make x or y negative,) the various numerical values of x and y become known.

The number of answers will be limited when the signs of p in the values of x and y are unlike; but unlimited when p has the same sign in both.

* If a and b have a common divisor, it must also be a divisor of c, or the problem is impossible.

† This expression is used to designate any whole number.

EXAMPLES.

1. Given 19x=14y+15, to find the values of x and y

in whole positive numbers.

If now, we assume p successively =0, 1, 2 3, &c.

y=3, 22, 41, 60, &c.

x=3, 17, 31, 45, &c.

Here the number of answers is evidently unlimited.

2. Given 11x+17y=987, to find x and y in whole positive numbers.

Whence y=11p+5,

And x-987-(11p+5)×17=82-172

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