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= 4.

Given 2+91 +42=80, to find 2.
Here 2 is the first figure of the root found by trial.

=9, b=4, N=80, r=2.
b = 4

9. Root
rrtc
r(r+c) = 22

80(2.47213596 First divisor = pi + cr + b 26

52 r. stu &c. 22

28 = N' gb + c's =s(s+ 3r+6)=

b 6.16 23.264 2d divisor is si + c's + b - 58.16 1.736 N'

som

. 16 4.593323 t(t +0+ 3(7 + 8)= 1.1389

.142677 *398 3d divisor is f + d't + '=65.6189

b

133591 334 And so on,

49

9086 64 (9+(2.47X3) +.002)2 = 16.4112

032824

6683 60 Fourth divisor is

66,79515

2403 4 16.3161 X1 • 116.4161 2

2005 Fifth divsior is.

616.81310

*398 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 fig. ures 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 2*, 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 ineluding 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 2 + 4x2 + 2x 2328, to find the value of x. Here the highest denomination of the root found by trial is 10. 6

4 2 2328 ( 12 root. 14....140, 142 Here it must be observed, as in quadratics Divisor 142 908 that the 1 in the quotient must be considered

100 908 10 thus, (rtar= (10+4) X10 = 140; 36.... 72 and so if were in the hundred's place it Divisor 454

must be considered as 100, &c.

Biquadratic Equations. For the solution of equations of a higher order, the principle is

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=Bx

= D.23 = Ex= N.

a

nearly the same. Let Ex*+Dr+Cr?+Ba=N be any biquadratic equation ; let r be the first figure in the root; then we shall

N have

Now let y denote the remaining Eri+Dr+Cr+B figures of the root, then ytr=t, and by substitution,

By + Br
C y + 2Cry + Cra Cx?
Dyu + 3 Dr y + 3Dry + Dr8

Dp
Ey + 4Erya + 6 Er'y + 4Ery + Er
Ey + D'yx + C' y+ B'y + A

Or, Ey4+D'y+C'y?+B'y=N', an equation similar to the one proposed ; and if S be the first figure of the root, then we shall

N' have

S. Now to find the denoES + D'S? + C'S + B' minators, put down the coefficients in a line, thus ; E D с

B Er+D, r+ Dr+C, r+ Dr+ Cr+B 2ErD, 3E7?12Dr+c 4Er+3Dv2+2Cr+B=B' 3Er+D, 6Er?+3Dr+C=C 4Er+ D=D', E' D'

C

B' Es + D', Es? + D's + C', Esø + D'sa + C's + B' 2Es + D', 3Es + 2D's + C', 4Es + 3D's: +2C's + B' 3Es + D', 6ES: + 3D's + C',

, 4Es + D, &c. &c.

&c. 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, Fz+Ex*+Dx+Cx?+Bx=N. Let r be the first figure of the root. . Then rty=t, or ytr=.

By + Br

Вх C aj + 2Cr y + Ci? Cx? Dy + 3 Dry

3 Dry

D703 Dx? Ey + 4Eryl + 6Eržy? + 4Ery + Er4 = Ex Fys + 5 Fryt + 10 Fry + 10 Fr?j? + 5 Fr*y + Fr5 = F2 Fyo + E'y + D'ya + C'y + B'y + A = N

Or, Fy*+E'y+D'yi+C'y?+B'y=N', an equation similar to the proposed one whose root s must be such, that if N' be divided by ** E$ + 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,

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# 3D2

D',

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17*

F
D

B
Fr E, F + Er + D, Fy8+ Er? + Dr + C; Fri Er8 + Dalt Cr + B
2 Fr E, 3Fp8 + 2Er

3E72 + 2Dr + C;5F74
3 Fr

C
E, 6F22 + 3Er + D, 107,8 + 6Er + 3Dr + C=C'
4.Fr E, 10Fr% + 4Er + D
5F,

E'.
F
E
D

B'
Fs + E Fg + E's + D' Fs' + E'si + D's + C Fs4 + E's + D'j? + C's + B'
2Fs + E 3F5 + 2E's
3Fs E 6Fs? + 3E's D' 10 F5P + 6E'se + 3D's + C'
4Fs + E' 10Fs? + 4E's + D'
5Fs + E',
&c.
&c.

&c.
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

Ez= determined. Lastly, let us take the general equation. Let Bx"+::.. Ex'+D.'+Cx' + Bx N 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

a

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= Ñ,

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series 0, 1, 2, 3, ; or 10, 20, 30, foc. ; that when it and the next succeeding number are separately substituted for a 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 Rp to.. Eri+Dr+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 ytrex, and therefore

By + Br = Bx,
Cy' + 2Cry + Cri = Cx",
Dy + 3 Dry + 3Dry + Dp = Dxo,
Ey + 4Ery: + 6Ery+ 4Ery+ Ert= Et,
Ry" +....
Ry" +. ...E'y+ D'ye + C'y + B'y

+A and, by transposition, Ry" + .... Ey+ D'ya + C'ye + B'y = N - A = N', an equation similar to the one proposed, the first figures, in the root y of which must be such, that if N'be divided by Rs-+

E's+D's? +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"??+C":+B'z=N", also similar to the first; and if we continue this process, we may obtain, one by one. all the fig. ures of the root 2, 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 Ryo+:...E'y+D'y+C'y?+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, fc.; 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, fc. respectively, then the first, second, third, &c. columns, without the numeral coefficients, will be R, Rr+Q, Rri+Qr+P, fc.; 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 fc. f.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.

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77 267*

find X.

Given av +902% + 2700x = Given 2+2=500, to find 2 to 21168, to find x. +90 10 or 12 places.

+1 2700 21168(6.384761

0 500(7.61727975 96.... 576 19656 [9678......56 392 [594 Div. 3276 1512

Div. 56 108 36 1176.147

49 104.736 108.3.. 32.49

335.853
22.6. 13.56

3.264
Div. 3920.49 316.943072 Div.174.56 1.887181

9
18.909928

36 1.376819 108.98..8.7184

15.883799
23.81..23.81

1.323862113 Div. 3961.7884

3.026129
Div. 188.7181

52956887
64
2.780024

1

37858967 109.1/44.. 4366

23.837.. 166859 246105

15097920 Div. 3970.9498 238292

Div. 189.123159

13251090

49 7813

1846830 Div. 3971.463 288

3972
23|85|12..4770

1703729 Div. 3|917|1.154 29

Div. 189.294837 3841

1124 143101

167 28 3574*

946 132512 Given 20+ 2.x— 3x=9, to Div. 189.30128

178 10589 +2

2

170 9465 -3

9(1.9394650535 Div.1,8,9.3,0,3,2 3..3 8.379

Given 28 +102 + 5x=260, to Div, 0 .621

+10
1
469857

5 260(4.11798686 5.9..5.31 151143

14.....56 244 Div. 9.31 143684019

Div. 61 81

16 13.521 7458981 7.73..2319

22.1.... 2.21 6415316

2.479 Div.15.6619

Div. 135.21 1.376531
1043665
9

1
962514

1.102469 7.799 .. 70191

2231

966091 81151 Div.15.964891

Div. 137.6531 80212

136378 81

1

124372 939 7.8/17...313

2/2.337 ... 1570 802

120 12006 Div.16.03829

Div.

138.013 111 11057
137
5

21
128

9 949

138.191
16.0419
9

8 829

1,3,8,.2,1 1,6.0,4,2,4 8

120 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.

find 2.

16

22.31.....

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