The sums or errors = 17.675 65x=26.80 -50=-50. 22.664 (22.664)-(17.675)=4.989. Hence (Art. 503.) 4.90.117.63. (4.1)-3=1.1, or 1 nearly. x=1, and x-1=0. x-1)x3-16x2 +65x--50(x2-15x+50 x3-x2 -15x2 +65x 50x-50 50x-50 x2-15x-50. 4x2-60x+225 +25. 2x-15=5. 2x =15+5=20 or 10. x=10 or 5. Therefore the roots of the equation are, 1, 5, 10. Example 4. What are the roots of the equation x3+2x2 -33x=90? By transposing 90, the equation becomes, x3+2x2-33x -90-0. Let x=6.1, or 6.2. Then by substituting these numbers for x in the given equation, we have, By the 1st supposition: The snms or errors = =10.101 By the 2d supposition: x3=238.328 2x276.88 -33x=-204.6 -90=-90 20.608 (20.608)-(10.101)=10.507= the difference of the errors. Hence, 10.507 0.1::10.101 0.1, the correction. This subtracted from the value assigned to x by the 1st supposition becomes (6.1)-0.1=6=x. x-6)x3-2x2-33x-90(x2+8x+15 23-6x2 8x2-33x 8x2-48x 15x-90 15x-90 -5 or x2+8x=-15. x2+8x+16=1. x+4=±1. x= -3. Therefore the roots of the equation are 6, —5, -3. REMARK. When the largest supposed number approxi mates nearest to the true value, the correction may be added; but if the smallest number approaches nearer, the correction may be subtracted. The two examples which immediately follow, are per. formed according to the previous remark. Example 5. What is a near value of the roots of the equation x3+9x2+4x=80? Let the roots=2.45 or 2.46. Then by substituting, we have, By the 1st supposition. x3=(2.45)3=14.706125 9x2 (2.45)2=54.0225 4x=4(2.45) = 9.80 -80. -1.471375 By the 2d supposition : x3=(2.46)3=14.886936 9x2(2.46)2=54.4644 4x=4(2.46) 9.84 -80. -0.808664 (-1.471375)-(-0.808664)=0-0.662711= the differ ence of the errors. Hence, (Art. 503.)0.660.1::0.80: 0.1, the correction. (2.46)+(0.1)=2.47, the answer. Example 6. What is a near value of one of the roots of the equation x3+x2+x=100? Let the roots 4.27 or 4.28. By substituting we have, By the 1st supposition: By the 2d supposition : x3 (2.48)3-78.402752 x2=(2.48)2=18.3184 x3=(2.47)3=77.854483 = 4.28 = 4.27 -100. x = 2.48 -100 (2.001152) the errors. (0.357383)=1.643769= the difference of Hence (Art. 503.) 1.64; 0.1::0.357 : 0.2, the correction, (4.27)-(0.2)=6.25. NEWTON'S METHOD OF APPROXIMATION. Four examples under article 503. b. Example 1. Find the values of x in the equation 3 16x265x50. Taking the reduced equation 2 = x=r-z=11-0.8-10.2. To obtain a nearer approximation to the root, let the corrected value be now substituted for r in the preceding equation, instead of the assumed value 50-1061.208+1664.64-663 11. We shall then have z = -312.12+326.4-65 = -9.568 -50.72 =0.188. As x=r-z, substituting figures, x= 10.2-0.188=10.012. For a third approximation, let the corrected value, 10.012, be substituted for r, and we then 50-1003.604322328+1603.842304-650.720 have z = .542018328 -45.336432 the former corrected value, we have z=10. Example 2. What is a near value of one of the roots of the equation, x3+10x2+5x=2600? By substituting r-z=x, we have (r-z)3+10(rz)3+5 (r-z)=2600. By expanding the equation becomes, r33r2z+3rz2 —z3+10r2—20rz+z2+5r-5z=2600. By can. celling and uniting, r3-3r2z+10r2-20rz+5r-5z=2600. Transposing, -5%-20rz-3r2z=2600-5r+10r2-r3. Di.. 2600-5r-10r2 —r2 viding, z= -5-20r-3r2 Assuming r 11, and 2600-55-1710-1331 substituting, z= -5-220-363 = = 4 -588 0.0068. Adding -0.0066 to 11, it becomes 11.0068=the answer within one ten thousandth. Example 3. What are the roots of the equations 3x+2x2 -11x=12? Substituting r―z for x, the equation becomes, (r-z)3 + 2(r-z)-11(r-z)=12. Expanding, r3-3r2z+3rz2 +z3 +2r2-4rz+2z2 —11r—11z=12. Uniting terms, 33raz+2r2-4rz-11r+11z=12. Transposing, 11x-4rz 3r2z=12—r3—2r2+11r. Dividing, z=. Assuming r=3, and substituting, z = 12-r2-2r2+11r 11-4r-3r2 12-27-18+33 11-12-27 0 =-0.3, nearly. As -0.3, is small, it may be reject -28 ed, and r be considered=3, the first root of the equation. To find the others, let x3+2x2-11x-12 be divided by x-3. x−3)x3 +2x2 —11x-12(x2+5x+4 5x2-11x 5x2-15x 4x-12 4x-12 In x2+5x+4, transpose and complete the square, 4x2 +20x +25=-16+25. Extracting the square root, 2x+5=±3, x=-2, or -8. Again, 2x+5=±3, and x=-4, or - 1. Therefore the roots of the equation are +3, -2 or 8. -4 or - 1. Example 4. What are the roots of the equation xa +4x3 -7x2-34x=24? By substituting r-z for x, the equation becomes, (r-z) 1+ 4(r-z)3-7(r−z)2-34 (r-z)=24. Expanding, r2-4r2z+6r3z2 -4rz3+24+4r3-12r2z+12rz2-423-7r2 +14rz-722-34r+ 34% 24. Uniting terms, r3-4r3z+4r3-12r2z-7r2+14rz34r+34x=24. Transp. -4r3z-12r2z+14rz+34x=24+ 34r+7r2-4r3_r4. Dividing, z = 24+34r+7r2 - 4r3_r4 −4r3-12r2+14r+34 Assuming r=3, and substituting, z= 24+102+63 - 108 - 81 -108 108+42+34 values of x=3, and x-3=0. 0 0. Therefore the 1st -140 x-3)x4+4x3-7x3-34x-24(x3+7x2+14x+8=0 Substituting r-z for x, in the above quotient, and the equation becomes, (r-z)3+7(r-z)2+14(-2)+8=0. Expanding, r3–3r2z+3rz2−z3+7r2−14rz+7z2+14r−14z+8=0. ting terms, r3-3r2z+7r2−14rz+14r-14z+8=0. Transposing, - 3r2z - 14rz – 14z=−8–14r-7r2-r3. Dividing, z= Uni -8-14r-7r2_p3 -3r2-14r-14 -8+56-112+64 Assuming r=-4, and substituting, z 0 0. Therefore the 2d value of a 6 -484-56-14 =-4. The others may be obtained by division. Dividing, x3+7x2+14x+8 by x+4, and the quotient becomes, x2 +3x +2. Transposing, x2+3x=-2. Completing the square, 4x2+12x+9=9-8. Extracting the square root, 2x+3= +1, and x=-2 or 1. Therefore the roots of the equation are – 1, − 2, − 4, +3. These may be proved to be the true roots, by involving them severally, and adding them together. From the known relations between the roots and co-efficients of equations, Newton has derived a method of determining the co-efficients, from the sum of the roots, the sum of their squares, the sum of their cubes, &c. and vice versa. S, is put for the sum of the roots, S, is put for the sum of the squares, S, for the sum of the cubes, &c. Under this article there is but the following unsolved example. 3 3 Example 2. Article 503. e. Required the terms of the biquadratic equation, in which S,=1, S2-39, S1=-89, and the product of all the roots after the signs are changed, is - 30. 28 A=-1. B=-(-1×1+39)-(-1+39)=- -19. C 2 =-(—19×1+39x-1-89)=-3(-19-39-89) = -147 +49. Therefore the required equation is x4x319x2+49x-30=0. APPLICATION OF ALGEBRA TO GEOMETRY. Remarks comprising a brief synopsis of what precedes the problems. In reference to this branch of the mathematics, it is too true that it has been superficially studied by many, if not neglected altogether. The student often at the outset embraces the notion that it is something very difficult, and not essentially important. Accordingly the shortest, and what is supposed, the easiest method is had recourse to in getting |