Posted by Salman on .
Use Newton's method to approximate a root of the equation (2 x^3 + 4 x + 4 =0) as follows.
Let (x_1 = 1\) be the initial approximation.
The second approximation (x_2) is ?
and the third approximation (x_3) is ?

Calculus 
Reiny,
First of all, why not reduce it to
x^3 + 2x + 2 = 0
Your iteration formula should be
x_{new = (2x^3  2)/(3x^2 + 2) so for a start of x=1 we get x1 = 1.333333 x2 = .91919 x3 = .78357 BTW, x5 = .770917 which is correct to 7 decimal places.} 
Calculus 
Reiny,
looks like I forgot to close my subscript code
The end should read:
Your iteration formula should be
xnew = (2x^3  2)/(3x^2 + 2)
so for a start of x=1
we get
x1 = 1.333333
x2 = .91919
x3 = .78357
BTW, x5 = .770917
which is correct to 7 decimal places. 
3 x  6x + 4 = 0 
rajkumar,
using by newton's method an approximate value the positive root