Find a root of an equation `f(x)=sqrt(12)` using Fixed Point Iteration method

Solution:
`x=sqrt(12)`

`:.x^2=12`

`:.x^2-12=0`

Here

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

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

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


`:.12-x^2=0`

Adding `10x` in both the sides, we get

`10x=12-x^2+10x`

`:.x=(12-x^2+10x)/10`

`:.phi(x)=(12-x^2+10x)/10`

`x_1 = phi(x_0) = phi(3.5) = 3.475`

`x_2 = phi(x_1) = phi(3.475) = 3.46744`

`x_3 = phi(x_2) = phi(3.46744) = 3.46513`

`x_4 = phi(x_3) = phi(3.46513) = 3.46442`

`x_5 = phi(x_4) = phi(3.46442) = 3.4642`


Approximate root of the equation `(12-x^2+10x)/10` using Iteration method is `3.4642` (After 5 iterations)

`n`	`x_0`	`x_1=phi(x_0)`	Update	Difference `|x_1-x_0|`
2	3.5	3.475	`x_0 = x_1`	0.025
3	3.475	3.46744	`x_0 = x_1`	0.00756
4	3.46744	3.46513	`x_0 = x_1`	0.00231
5	3.46513	3.46442	`x_0 = x_1`	0.00071
6	3.46442	3.4642	`x_0 = x_1`	0.00022

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