= 4. Given 2+91 +42=80, to find 2. =9, b=4, N=80, r=2. 9. Root 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 ; =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 Dp 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, Ꭼr+ Dr+C, Ꭼr+ 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, 2 # 3D2 D', HHHHH 17* F B 3E72 + 2Dr + C;5F74 C E'. B' &c. ; 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 = Ñ, 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, +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. 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 3.264 9 36 1.376819 108.98..8.7184 15.883799 1.323862113 Div. 3961.7884 3.026129 52956887 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 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 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 1 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 21 9 949 138.191 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..... |