Question

find the real root of the equation f(x)=x^3+x^3+x+7=0 bisection method correct upto three decimal places​

Answer 1

Answer:

Let , f(x)=0 and two real numbers a and b such that f(a)*f(b)<0

Set c = (a+b)/2

if f(c)<0 then b=c, otherwise c=a

Repeat step 2 and 3 until (a-b)<DOA where DOA means Degree of accuracy.

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)

Hence root exits between x1=0.6796 and x2=0.6835

9th approximate root :

x9 = (a + b)/2 = (0.6796+0.6835)/2 =0.6816

f(x1) =0.6816^3+0.0.6816-1 = -0.0016(-ve)

Hence root exits between x1=0.6816 and x2=0.6835

10th approximate root :

x10 = (a + b)/2 = (0.6816+0.6835)/2 =0.6826

Hence the approximate correct upto 3 decimal places. So the real root of the equation = 0.6826