`(x-2)^2=ln(x)`

`f(x)=(x-2)^2-ln(x)=0`

`f'(x)=2(x-2)-1/x`

See the attached graph. The curve of the function intersects the x-axis at x `~~` 1.4 and 3. These can be used as initial approximates. Iterate until six digit decimal places are same.

`x_(n+1)=x_n-((x_n-2)^2-ln(x_n))/(2(x_n-2)-1/x_n)`

For x_1=1.4

`x_2=1.4-((1.4-2)^2-ln(1.4))/(2(1.4-2)-1/1.4)`

`x_2~~1.412291`

`x_3~~1.412391`

`x_4~~1.412391`

For x_1=3

`x_2=3-((3-2)^2-ln(3))/(2(3-2)-1/3)`

`x_2~~3.059167`

`x_3~~3.057106`

`x_4~~3.057103`

`x_5~~3.057103`

Roots of the equation are 1.412391 , 3.057103

Images:

This image has been Flagged as inappropriate Click to unflag

Image (1 of 1)