First, It will be found that x is more than 4, and less than 5. Substituting these numbers for x, we have x3 10x2 64 160 20 244 68.921 168.10 20.5 257.521 1 -15 63 49 400 Then, 400-244: 5-4 :: 260-244: the correction . 1. This correction added to the less assumed number, gives 4.1 for an approximate value of x. Secondly, By substituting 4.1 and 4.2 for x, we have x3 10x2 -15.9135 64.89 50.069227 First, Substituting 1 and 1.5 for x, x3 -15x2. 63x -15.x2 63x 125 · 250 25 Then, 271.488-257.521: 4.2-4.1 260-257.521: the correc tion .017. Adding this correction to the less assumed number 4.1, we have 4.117 for a nearer value of x. 7. Find an approximate value of x in the Equation x3-15x2+63x=50. Then, 64.125-49: 1.5—1:50-49: the correction .03. 1.092727 . x3. 3.375 -33.75 94.5 64.125. 1.061208 -15.606 64.26 49.715208. Then, 50.069227-49.715208: 1.03-1.02: 50-49.715208: the "ection .00804. rlence x 1.02+.008041.02804'. 8. Find an approximate value of x in the Equation x3-17x2+54x=350. -3799.5425 807.30 349.119875 -17x2 54x 10000. -300 -750 8950 Then, 360-168:15-14:: 360-350: the correction .05. Subtracting this correction from the greater assumed member, we have 14.95 for an approximate value of x. Secondly, Substituting 14.95 and 14.96 for x, we have 3341.362375 . x3 -17x2 54.x 9. Find an approximate value of x in the Equation x4-3x2-75x=10000. First, Substituting 10 and 11 for x, we have x4 -3x2 752 X4 -3.x2 -75x 807.84 351.284736. Then, 351.284736-349.119875: 14.96-14.95: 350-349.119875: the correction .004065. Adding this correction to the less assumed number 14.95, we find 14.954065 for a very near value of x. 3375 -3825 810 360. -3804.6272 3348.071936 Then, 13453-8950: 11-10 :: 10000-8950: the correction .23. Hence 10.23 is the first approximation to the value of x. Secondly, Substituting 10.2 and 10.3 for x, 10824.3216 -312.12 -765 9747.2016 14641 -363 -825 13453. 11255.0881 -318.27 -772.5 10164.3181. Then, 10164.3181-9747.2016: 10.3-10.2:: 10164.3181-10000: the correction .039393. Hence 10.3-.039393=10.260607, is a nearer value of x. 10. Find an approximate value of x in the Equation 2x4-16x3+40x2-30x+1=0. -39 1 --35.152 67.6 1 3 25 .1602 -30x 31.622 56.622 Then, 5-(-3): 2-1::0-(-3): the correction .3. Secondly, Substituting 1.3 and 1.2 for x, 5.7122 2x4 -16x3 40.x2 -30.x 1 ( Substituting 10 and 11 for x, 11. Find an approximate value of x in the Equation 2 x2-15) +x√x=90. -15)2 + -.9008. Then, .1602-(-9008): 1.3-1.2:: 0-(-9008): the correction .084. Hence 1.2+.084–1.284, is a nearer value of x. 1 5. -27.648 57.6 -36 1 4.1472 84.64 36.482 121.122. Then, 121.122-56.622:11-10 :: 121.122-90: the correction .48. Therefore 11.48 10.52, is an approximate value of x. PROBLEMS IN INEQUATIONS. 1. Find a number such, that when multiplied by 5, the product shall be less than 40, and when 5 is added to 3 times the square of the number, the sum shall be greater than 80. Let x represent the number; then we shall have 5x40, and 3x2+5>80. Dividing the first Inequation by the coefficient of x, we find x<8. Transposing 5 in the second Inequation, and adding similar terms, 3x2>75; hence x2>25, and x>5. The required number is therefore indeterminate within the limits 5 and 8; that is, it is either of the numbers 5, 6, cr 7, plus any proper fraction. 2. What number is that whose third part diminished by 3, is greater than 20, and whose fourth part increased by 4, is less than 30? Let x represent the number; then we have Xx 3 from which we shall find x>69, and <104. X -320, and 4<30; 4 3. Find a number whose square diminished by 10 is less than 90, and whose square root increased by 2 is greater than 5. Let x represent the number; then x2-10<90; and √x+2>5. These Inequations will give x<10, and >9; hence the number is 9, plus any proper fraction. 4. Find a number such, that if it be multiplied by 2, 3, and 4, successively, the sum of the products shall be greater than 100, and if it be divided by the same numbers, the sum of the quotients shall be less than 30. Let x represent the number; then X X x 2x+3x+4x100; and + + <30; 2 3 from which we shall find-x>114, and <27 5. A farmer sold a number of cattle. If 6 be subtracted from 3 times the number, the remainder will be less than the number increased by 48; and if 6 be added to 4 times the number, the sum will be greater than 27 increased by 3 times the number. What was the number of cattle? Let x represent the number; then 3x-6x+48; and 4x+6>27+3x; from which x<27 and >21; hence the number of cattle was 22, 23, 24, 25, or 26. 6. Find a number whose square added to 4 times the number itself shall be more than 45, but whose square diminished by 10 times the number shall be less than 75. Let x represent the number; then 4x+x245; and x2-10x<75; Completing the square in the first member of each of these Inequations, we have x2+4x+4>49; and x2-10x+25<100. Extracting the square root of each member, from which x>5 and <15. 7. Five times the number of miles between two places increased by 10, is less than 100; and 8 times the number diminished by 5, is greater than the number increased by 100. Required the number of miles between the two places. Let x represent the number of miles; then 5x+10100; and 8x-5x+100. These Inequations will give x<18 and >15; therefore the number of miles was 16, or 17, or either 15, 16, or 17, plus any proper fraction. |