Question

Find a real root of x
3 − x = 1 between 1 and 2 by bisection method. Compute correct to five decimal
places.

Answer 1

Answer:

The given equation is f(x)=x^3+x-1=0

Let a=0, b=1

First approximate root :

x1 = (a + b)/2 = (0+1)/2 = 0.5

f(x1) = 0.5^3+0.5-1 = -0.375 (-ve)

Hence root exits between x1=0.5 and x2=1.0

Second approximate root :

x2 = (a + b)/2 = (0.5+1)/2 =0.75

f(x1) = 0.75^3+0.75-1 = +0.17 (+ve)

Hence root exits between x1=0.5 and x2=0.75

Third approximate root :

x3 = (a + b)/2 = (0.5+0.75)/2 =0.625

f(x1) = 0.625^3+0.625-1 = -0.130 (-ve)

Hence root exits between x1=0.625 and x2=0.75

Fourth approximate root :

x4 = (a + b)/2 = (0.625+0.75)/2 =0.6875

f(x1) =0.6875^3+0.6875-1 = +0.012 (+ve)

Hence root exits between x1=0.625 and x2=0.6875

Fourth approximate root :

x5 = (a + b)/2 = (0.625+0.6875)/2 =0.6562

f(x1) =0.6562^3+0.6562-1 = -0.0611 (-ve)

Hence root exits between x1=0.6562 and x2=0.6875

Sixth approximate root :

x6 = (a + b)/2 = (0.6562+0.6875)/2 =0.6718

f(x1) =0.6718^3+0.6718-1 = -0.0248(-ve)

Hence root exits between x1=0.6718 and x2=0.6875

Seventh approximate root :

x7 = (a + b)/2 = (0.6718+0.6875)/2 =0.6796

f(x1) =0.6796^3+0.6796-1 = -0.0063(-ve)

Hence root exits between x1=0.6796 and x2=0.6875

8th approximate root :

x8 = (a + b)/2 = (0.6796+0.6875)/2 =0.6835

f(x1) =0.6835^3+0.6835-1 = +0.0030(+ve)