Find `sqrt(12)` using Secant 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 :

`x_0 = 3` and `x_1 = 4`

`f(x_0) = f(3) = -3` and `f(x_1) = f(4) = 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`


`2^(nd)` iteration :

`x_1 = 4` and `x_2 = 3.4286`

`f(x_1) = f(4) = 4` and `f(x_2) = f(3.4286) = -0.2449`

`:. x_3 = x_1 - f(x_1) * (x_2 - x_1)/(f(x_2) - f(x_1))`

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

`x_3 = 3.4615`

`:. f(x_3)=f(3.4615)=3.4615^2-12=-0.0178`


`3^(rd)` iteration :

`x_2 = 3.4286` and `x_3 = 3.4615`

`f(x_2) = f(3.4286) = -0.2449` and `f(x_3) = f(3.4615) = -0.0178`

`:. x_4 = x_2 - f(x_2) * (x_3 - x_2)/(f(x_3) - f(x_2))`

`x_4 = 3.4286 - (-0.2449) * (3.4615 - 3.4286)/(-0.0178 - (-0.2449))`

`x_4 = 3.4641`

`:. f(x_4)=f(3.4641)=3.4641^2-12=0.0001`


Approximate root of the equation `x^2-12=0` using Secant method is `3.4641` (After 3 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_1` `x_1 = x_2`
2	4	4	3.4286	-0.2449	3.4615	-0.0178	`x_0 = x_1` `x_1 = x_2`
3	3.4286	-0.2449	3.4615	-0.0178	3.4641	0.0001	`x_0 = x_1` `x_1 = x_2`

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