Find `sqrt(12)` using Bisection method

Solution:
Let `x=sqrt(12)`

`:.x^2=12`

`:.x^2-12=0`

i.e. `f(x)=x^2-12`

Here

`x`	0	1	2	3	4
`f(x)`	-12	-11	-8	-3	4

`1^(st)` iteration :

Here `f(3) = -3 < 0` and `f(4) = 4 > 0`

`:.` Now, Root lies between `3` and `4`

`x_0=(3+4)/2=3.5`

`f(x_0)=f(3.5)=3.5^2-12=0.25 > 0`


`2^(nd)` iteration :

Here `f(3) = -3 < 0` and `f(3.5) = 0.25 > 0`

`:.` Now, Root lies between `3` and `3.5`

`x_1=(3+3.5)/2=3.25`

`f(x_1)=f(3.25)=3.25^2-12=-1.4375 < 0`


`3^(rd)` iteration :

Here `f(3.25) = -1.4375 < 0` and `f(3.5) = 0.25 > 0`

`:.` Now, Root lies between `3.25` and `3.5`

`x_2=(3.25+3.5)/2=3.375`

`f(x_2)=f(3.375)=3.375^2-12=-0.6094 < 0`


`4^(th)` iteration :

Here `f(3.375) = -0.6094 < 0` and `f(3.5) = 0.25 > 0`

`:.` Now, Root lies between `3.375` and `3.5`

`x_3=(3.375+3.5)/2=3.4375`

`f(x_3)=f(3.4375)=3.4375^2-12=-0.1836 < 0`


`5^(th)` iteration :

Here `f(3.4375) = -0.1836 < 0` and `f(3.5) = 0.25 > 0`

`:.` Now, Root lies between `3.4375` and `3.5`

`x_4=(3.4375+3.5)/2=3.4688`

`f(x_4)=f(3.4688)=3.4688^2-12=0.0322 > 0`


`6^(th)` iteration :

Here `f(3.4375) = -0.1836 < 0` and `f(3.4688) = 0.0322 > 0`

`:.` Now, Root lies between `3.4375` and `3.4688`

`x_5=(3.4375+3.4688)/2=3.4531`

`f(x_5)=f(3.4531)=3.4531^2-12=-0.0759 < 0`


`7^(th)` iteration :

Here `f(3.4531) = -0.0759 < 0` and `f(3.4688) = 0.0322 > 0`

`:.` Now, Root lies between `3.4531` and `3.4688`

`x_6=(3.4531+3.4688)/2=3.4609`

`f(x_6)=f(3.4609)=3.4609^2-12=-0.0219 < 0`


`8^(th)` iteration :

Here `f(3.4609) = -0.0219 < 0` and `f(3.4688) = 0.0322 > 0`

`:.` Now, Root lies between `3.4609` and `3.4688`

`x_7=(3.4609+3.4688)/2=3.4648`

`f(x_7)=f(3.4648)=3.4648^2-12=0.0051 > 0`


`9^(th)` iteration :

Here `f(3.4609) = -0.0219 < 0` and `f(3.4648) = 0.0051 > 0`

`:.` Now, Root lies between `3.4609` and `3.4648`

`x_8=(3.4609+3.4648)/2=3.4629`

`f(x_8)=f(3.4629)=3.4629^2-12=-0.0084 < 0`


`10^(th)` iteration :

Here `f(3.4629) = -0.0084 < 0` and `f(3.4648) = 0.0051 > 0`

`:.` Now, Root lies between `3.4629` and `3.4648`

`x_9=(3.4629+3.4648)/2=3.4639`

`f(x_9)=f(3.4639)=3.4639^2-12=-0.0016 < 0`


`11^(th)` iteration :

Here `f(3.4639) = -0.0016 < 0` and `f(3.4648) = 0.0051 > 0`

`:.` Now, Root lies between `3.4639` and `3.4648`

`x_10=(3.4639+3.4648)/2=3.4644`

`f(x_10)=f(3.4644)=3.4644^2-12=0.0018 > 0`


`12^(th)` iteration :

Here `f(3.4639) = -0.0016 < 0` and `f(3.4644) = 0.0018 > 0`

`:.` Now, Root lies between `3.4639` and `3.4644`

`x_11=(3.4639+3.4644)/2=3.4641`

`f(x_11)=f(3.4641)=3.4641^2-12=0.0001 > 0`


Approximate root of the equation `x^2-12=0` using Bisection method is `3.4641` (After 12 iterations)

`n`	`a`	`f(a)`	`b`	`f(b)`	`c=(a + b)/2`	`f(c)`	Update
1	3	-3	4	4	3.5	0.25	`b = c`
2	3	-3	3.5	0.25	3.25	-1.4375	`a = c`
3	3.25	-1.4375	3.5	0.25	3.375	-0.6094	`a = c`
4	3.375	-0.6094	3.5	0.25	3.4375	-0.1836	`a = c`
5	3.4375	-0.1836	3.5	0.25	3.4688	0.0322	`b = c`
6	3.4375	-0.1836	3.4688	0.0322	3.4531	-0.0759	`a = c`
7	3.4531	-0.0759	3.4688	0.0322	3.4609	-0.0219	`a = c`
8	3.4609	-0.0219	3.4688	0.0322	3.4648	0.0051	`b = c`
9	3.4609	-0.0219	3.4648	0.0051	3.4629	-0.0084	`a = c`
10	3.4629	-0.0084	3.4648	0.0051	3.4639	-0.0016	`a = c`
11	3.4639	-0.0016	3.4648	0.0051	3.4644	0.0018	`b = c`
12	3.4639	-0.0016	3.4644	0.0018	3.4641	0.0001	`b = c`

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