Posted by **Salman** on Wednesday, November 11, 2009 at 4:31am.

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**, Wednesday, November 11, 2009 at 8:14am
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.
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- Calculus -
**Reiny**, Wednesday, November 11, 2009 at 1:21pm
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**, Wednesday, October 21, 2015 at 10:26am
using by newton's method an approximate value the positive root

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