Math, asked by sivathmikad, 6 months ago

A root of the equation x3 - x - 11 = 0 using bisection method is​

Answers

Answered by amitnrw
10

Given : x³ - x - 11  = 0

To Find : a  root  by bisection method

Solution:

x³ - x - 11  = 0

f(0)  = - 11

f(1)  =  - 11

f(2) = -5

f(3)  = 13

one root lies between 2 & 3

(2 +3 )/2 = 2.5

f(2.5) = 2.125

0 lies  between  -5 & 2.125

Hence (2 + 2.5)/2 = 2.25

f(2.25)  =  -1.859375

(2.25 + 2.5)/2  = 2.375

f(2.375) = 0.021484375

f(2.3125) = -0.946044922

f(2.34375)= -0.469146729

f (2.359375) = -0.225559235

f(2.3671875)= -0.102470875

f(2.37109375) -0.04060179

f(2.373046875) =  -0.009585865

f(2.374023438)=  0.005942463

f(2.373535156) =  -0.001823399

f(2.373779297) =  0.002059108

f(2.373657227)= 0.000117748

f(2.373596191 ) = -0.000852852

f(2.373626709) =  -0.000367558

f(2.373641968) -0.000124907

f(2.373649597) -3.57949E-06

f(2.373653412)  5.70844E-05

f(2.373651505) 2.67524E-05

2.37365  is a root

Learn More:

Use bisection method to find root of the equation x3 – 2x – 5 = 0 ...

https://brainly.in/question/6242554

Find the root of the equation x^3-2x-5= 0 by Bisection method ...

https://brainly.in/question/18811634

Similar questions