Find the root of x^3-9x+1=0
equation by using Bisection method, correct to one place of decimal in the interval (2, 4).
Answers
Given : x^3-9x+1=0
To Find : root using Bisection method, correct to one place of decimal in the interval (2, 4).
Solution:
x³-9x+1=0
x =2 => 2³ - 9(2) + 1 = -9
x = 4 => 4³ - 9(4) + 1 = 29
(2 + 4)/2 = 3
x = 3 => 3³ - 9(3) + 1 = 1
Hence lies between 2 and 3
(2 + 3)/2 = 2.5
and so on
x f(x)
2.5 -5.875
2.75 -2.953125
2.875 -1.111328125
2.9375 -0.090087891
2.96875 0.446258545
2.953125 0.175922394
2.9453125 0.042377949
2.94140625 -0.023989618
2.943359375 0.009160481
2.942382813 -0.007422986
2.942871094 0.000866643
2.94
and 2.9 upto one decimal place
2.9 is root of x³-9x+1=0 equation by using Bisection method, correct to one place of decimal in the interval (2, 4).
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Answer:
Find the root of x^3-9x+1=0
equation by using Bisection method, correct to one place of decimal in the interval (2, 3).