Find `sqrt(12)` using False Position method (regula falsi 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 `x_0 = 3` and `x_1 = 4`

`x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_2=3 - (-3) * (4 - 3)/(4 - (-3))`

`x_2=3.4286`

`f(x_2)=f(3.4286)=3.4286^2-12=-0.2449 < 0`


`2^(nd)` iteration :

Here `f(3.4286) = -0.2449 < 0` and `f(4) = 4 > 0`

`:.` Now, Root lies between `x_0 = 3.4286` and `x_1 = 4`

`x_3 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_3=3.4286 - (-0.2449) * (4 - 3.4286)/(4 - (-0.2449))`

`x_3=3.4615`

`f(x_3)=f(3.4615)=3.4615^2-12=-0.0178 < 0`


`3^(rd)` iteration :

Here `f(3.4615) = -0.0178 < 0` and `f(4) = 4 > 0`

`:.` Now, Root lies between `x_0 = 3.4615` and `x_1 = 4`

`x_4 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_4=3.4615 - (-0.0178) * (4 - 3.4615)/(4 - (-0.0178))`

`x_4=3.4639`

`f(x_4)=f(3.4639)=3.4639^2-12=-0.0013 < 0`


`4^(th)` iteration :

Here `f(3.4639) = -0.0013 < 0` and `f(4) = 4 > 0`

`:.` Now, Root lies between `x_0 = 3.4639` and `x_1 = 4`

`x_5 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_5=3.4639 - (-0.0013) * (4 - 3.4639)/(4 - (-0.0013))`

`x_5=3.4641`

`f(x_5)=f(3.4641)=3.4641^2-12=0 < 0`


Approximate root of the equation `x^2-12=0` using False Position method is `3.4641` (After 4 iterations)

`n`	`x_0`	`f(x_0)`	`x_1`	`f(x_1)`	`x_2`	`f(x_2)`	Update
1	3	-3	4	4	3.4286	-0.2449	`x_0 = x_2`
2	3.4286	-0.2449	4	4	3.4615	-0.0178	`x_0 = x_2`
3	3.4615	-0.0178	4	4	3.4639	-0.0013	`x_0 = x_2`
4	3.4639	-0.0013	4	4	3.4641	0	`x_0 = x_2`

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