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Method and examples
Method
root of an equation using Bisection method
f(x) =
Find Any Root Root Between and

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Bisection method calculator to find a real root an equation
1. f(x) = 2x^3-2x-5
2. f(x) = x^3-x-1
3. f(x) = x^3+2x^2+x-1
4. f(x) = x^3-2x-5
5. f(x) = x^3-x+1
6. f(x) = cos(x)
7. f(x) = 2*cos(x)-x
8. f(x) = 2^x-x-1.7


Example
1. Find a root of an equation f(x)=x3-x-1 using Bisection method

Solution:
Here x3-x-1=0

Let f(x)=x3-x-1

Here
x012
f(x)-1-15



1st iteration :

Here f(1)=-1<0 and f(2)=5>0

:. Now, Root lies between 1 and 2

x_0 = (1 + 2)/2 = 1.5

f(x_0) = f(1.5) = 0.875 > 0


2^(nd) iteration :

Here f(1) = -1 < 0 and f(1.5) = 0.875 > 0

:. Now, Root lies between 1 and 1.5

x_1 = (1 + 1.5)/2 = 1.25

f(x_1) = f(1.25) = -0.29688 < 0


3^(rd) iteration :

Here f(1.25) = -0.29688 < 0 and f(1.5) = 0.875 > 0

:. Now, Root lies between 1.25 and 1.5

x_2 = (1.25 + 1.5)/2 = 1.375

f(x_2) = f(1.375) = 0.22461 > 0


4^(th) iteration :

Here f(1.25) = -0.29688 < 0 and f(1.375) = 0.22461 > 0

:. Now, Root lies between 1.25 and 1.375

x_3 = (1.25 + 1.375)/2 = 1.3125

f(x_3) = f(1.3125) = -0.05151 < 0


5^(th) iteration :

Here f(1.3125) = -0.05151 < 0 and f(1.375) = 0.22461 > 0

:. Now, Root lies between 1.3125 and 1.375

x_4 = (1.3125 + 1.375)/2 = 1.34375

f(x_4) = f(1.34375) = 0.08261 > 0


6^(th) iteration :

Here f(1.3125) = -0.05151 < 0 and f(1.34375) = 0.08261 > 0

:. Now, Root lies between 1.3125 and 1.34375

x_5 = (1.3125 + 1.34375)/2 = 1.32812

f(x_5) = f(1.32812) = 0.01458 > 0


7^(th) iteration :

Here f(1.3125) = -0.05151 < 0 and f(1.32812) = 0.01458 > 0

:. Now, Root lies between 1.3125 and 1.32812

x_6 = (1.3125 + 1.32812)/2 = 1.32031

f(x_6) = f(1.32031) = -0.01871 < 0


8^(th) iteration :

Here f(1.32031) = -0.01871 < 0 and f(1.32812) = 0.01458 > 0

:. Now, Root lies between 1.32031 and 1.32812

x_7 = (1.32031 + 1.32812)/2 = 1.32422

f(x_7) = f(1.32422) = -0.00213 < 0


9^(th) iteration :

Here f(1.32422) = -0.00213 < 0 and f(1.32812) = 0.01458 > 0

:. Now, Root lies between 1.32422 and 1.32812

x_8 = (1.32422 + 1.32812)/2 = 1.32617

f(x_8) = f(1.32617) = 0.00621 > 0


10^(th) iteration :

Here f(1.32422) = -0.00213 < 0 and f(1.32617) = 0.00621 > 0

:. Now, Root lies between 1.32422 and 1.32617

x_9 = (1.32422 + 1.32617)/2 = 1.3252

f(x_9) = f(1.3252) = 0.00204 > 0


11^(th) iteration :

Here f(1.32422) = -0.00213 < 0 and f(1.3252) = 0.00204 > 0

:. Now, Root lies between 1.32422 and 1.3252

x_10 = (1.32422 + 1.3252)/2 = 1.32471

f(x_10) = f(1.32471) = -0.00005 < 0


Approximate root of the equation x^3-x-1=0 using Bisection method is 1.32471

naf(a)bf(b)c=(a + b)/2f(c)Update
11-1251.50.875b = c
21-11.50.8751.25-0.29688a = c
31.25-0.296881.50.8751.3750.22461b = c
41.25-0.296881.3750.224611.3125-0.05151a = c
51.3125-0.051511.3750.224611.343750.08261b = c
61.3125-0.051511.343750.082611.328120.01458b = c
71.3125-0.051511.328120.014581.32031-0.01871a = c
81.32031-0.018711.328120.014581.32422-0.00213a = c
91.32422-0.002131.328120.014581.326170.00621b = c
101.32422-0.002131.326170.006211.32520.00204b = c
111.32422-0.002131.32520.002041.32471-0.00005a = c




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