Given x+9x+4x-80, to find z.

Here 2 is the first figure of the root found by trial. c9, b=4, N=80, r2.

1

Root

6.16

80(2.47213596 52 r. stu &c.

28 = N'

23.264

4.736

4.593323

N'

t(t + c + 3(r+s) = 1.1389

3d divisor is + c't + b' And so on,

(9+(2.47x3)+.002)216-412

.142677 *398

"

65.6189

49

.032824

Hence, from these formulas, we derive the following rule. Put down b the coefficient of x, and a little to the right place the absolute number, which is to be considered as a dividend, the figures of the root forming the quotient. Place the first figure of the root found by trial in the quotient, above which write the coefficient of 22, observing that its unit's place be over the unit's place of the quotient. Multiply the value of the quotient figure, taking in those above by that value; add the product to b, and the sum is the first divisor. Write the square of the quotient figure just found under the first divisor. Find now the next figure of the root, and to its value (including those above it) prefix three times the preceding, taking in the value of the figure above it. Multiply the result by the last found figure, add the product to the three sums immediately above, and we shall have the second divisor; and in the same manner are the succeeding divisors to be obtained.

=

Given 34x2 + 2x 2328, to find the value of x. Here the highest denomination of the root found by trial is 10. 4 = C

Here it must be observed, as in quadratics that the 1 in the quotient must be considered 10 thus, ( r+a)r = (10+4) X10 = 140; and so if were in the hundred's place it must be considered as 100, &c.

Biquadratic Equations.

For the solution of equations of a higher order, the principle is

nearly the same. Let Ex+Dx2+Cx2+B=N be any biquadratic equation; let r be the first figure in the root; then we shall

figures of the root, then y+r=x, and by substitution,

By Br

=Bx

Cy2+2Cry

Cr2

=Cx2

Dy3+3Dry + 3Dr2y + Dr3
y2+

Ey +4Ery3 + 6Er2y2 + 4Er3y + Era

= N.

Ey+ D'y3 + C' y2 + B' y + A Or, Ey1+D'y3+C'y2+B'y=N', an equation similar to the one proposed; and if S be the first figure of the root, then we shall

have

N

= S. Now to find the deno

ES3 + D'S2 + C'S + B' minators, put down the coefficients in a line, thus ;

Hence it is evident how these lines are formed, viz. by multiplying the succeeding terms by r, s, &c. and adding them respectively. By this means figurative numbers of the 2nd, 3d, &c. degrees are formed; (see Simpson's Alg.) and so we may proceed to an equation of any degree. Thus, let there be given the equation, Fx+Ex+Dx3+Cx2+Bx-N. Let r be the first figure of the root. Then r+y=x, or y+r=x.

=Bx

Cx2

By + Br
Cy2+2Cry + C12
Dy3+3Dry + 3Dr2y
3Dry Dr3 Dr3

Ex1

Ey 4 Ery3 6Ery2+4Er3y + Er1 Fy5Fry10 Fry+10 Fr3y2+5Fry + Fr5 = Fx5

Fy + E'y' + D'y3 + C'y+B'y + A = N Or, Fy3+E'y2+D'y3+C'y2+B'y=N', an equation similar to the proposed one whose root s must be such, that if N' be divided by Fs+ E's+D's+ C's+ B', the quotient will be s. Now, to find the denominator of these fractions, multiply by r, s, &c., as before. Thus,

D

Fr

E,

2 Fr
3 Fr
4Fr

E,

5 Fr

E, 10 Fr2
E=E'.

4 Er

D= D',

Fr2 + Er + D, 3F2Er D, 4F3.

C
Er2 + Dr + C; Fr
3Er2+2Dr+ C;5Fr1
E, 6F2 3Er D, 10Fr3+6Er2 + 3Dr + C

=C'

E'

D'

E' Fs2+ E's+ D' 3F2E's D'

5Fs E',

B'

Fs3+ E's2 + D's + C' Fs + E's3 + D's2 + C's + B'
4Fs34E's 2D's C' 5Fs*
D' 10 Fs3 6E's+3D's+ C'

FAE's 3D's 2C's + B'= B'

From these principles being found to hold good in all equations whatever, we can determine the following general rule. RULE. Arrange the coefficients of the given equation in a row, commencing with that of the first term.

Add the product of the first root figure, found by trial, and the first coefficient to the second coeffi-
cient; the product of the sum and same figure to the third coefficient, and so on to the last coefficient,
and the last sum will be the divisor. Repeat this process with the first coefficient and these sums, and
the number under the last sum will be the trial divisor for the next figure. Perform a similar process
with the first coefficient and these second sums, stopping under the n 1th coefficient. Perform
again a similar process with the same first coefficient and these laat sums, stopping here under the
preceding, or n 2th coefficient, and so on till the last sum falls under the second coefficient. Find
now from the trial divisor the next figure of the root, and proceed with the last set of sums, and this
new figure exactly the same as with the original coefficients and the first figure in.finding the prece-
ding divisor, and the next divisor will be obtained; and in a similar manner are the other divisors to be
determined. Lastly, let us take the general equation. Let Ba"....Ex2 + Dx3 + Cx2 + BN be an
equation of any degree whatever, and let r be such a number in either of the series 0, 1, 2, 3, &c.; or

series 0, 1, 2, 3, ; or 10, 20, 30, &c. ; that when it and the next succeeding number are separately substituted for x in the equation, the results shall be the one less, and the other greater than N; then r will be the first figure of one of the roots of the equation; and if N be divided by Rr... Er3+Dr2+Cr+B, the quotient must be r. Suppose such a value of r is found, and let y represent the succeeding figures of the root; then y+rx, and therefore By Br Bx,

Ey* + 4Ery3 + 6Er2y+ 4Er3y + Era— Exa,

N,

=

Ry+........ Ry+..........E'y'+ D'y3 + C'y2 + B'y + A = Ñ‚ and, by transposition, Ry....Ey+ D'y3 + C'y2 + B'y = N — A — N', an equation similar to the one proposed, the first figure s, in the root y of which must be such, that if N' be divided by Rs1+ ....E's3+D's2+C's+B′, the quotient will be s; if, therefore, we suppose s to be found, and if z be put for the remaining figures of the root, we shall, by proceeding as before, get another equation Rz"+E"z3+C"z+B"z= N"; also similar to the first; and if we continue this process, we may obtain, one by one. all the figures of the root x, and it is evident that each divisor will be similarly formed from the coefficients of the corresponding equation, and the new figure of the root.

Now it is obvious that B'x is the nth term in the equation Ry"....E'y'+D'y3+C'y2+B'y-N', or that the column represented by B' is the nth column from the left, and that it consists of n terms; the column represented by C' is the n 1th, and consists of n-1 terms, &c.; also each of these columns, omitting the numeral parts, is equal to the preceding multiplied by r, plus the corresponding coefficient in the proposed equation; consequently, since the first column is simply R, if the second, third, &c. coefficients of the proposed equation be Q, P, &c. respectively, then the first, second, third, &c. columns, without the numeral coefficients, will be R, Rr+Q, Rr2+Qr+P, &c.; but if the first term in this series be multiplied by r, and the product added to the second, and the result be multiplied by r, and the product added to the third &c. &c. that the proper numeral coefficients will be obtained, because the above columns of numeral coefficients are the same as the several binomial columns, except that the above are written in a reverse order, that is, from right to left, and in this reverse order they will be produced by adding the series, as above. See Emerson Algebra Ed. 1764 page 287.

28 3574*

946 132512

=

find x.

+2

-3

Given 3+2x2-3x-9, to Div. 189.30128

9(1.9394650585 Div.1,8,9.3,0,3,2

178 10589

2

170 9465

3..3

8.379

Given

+10+5x=260, to

+10

5

260(4.11798686

5.9..5.31

151143

A and B between them owe $240. A pays eight dollars a day, and B pays the first day $1, the second 2, the third 3, and so on. In how many days will they clear the debt, and how much did each of them owe? Ans. 15 days.