Solving a cubic equation.

**density** · September 23rd 2010, 10:59 AM

I have $f(x)= x^4 + 2x^2$

and am trying to solve

$f'(x)= 1$ , that is $4x^3 + 4x= 1$

**Unknown008** · September 23rd 2010, 11:03 AM

The common factor is 4x. Can you try to factor it?

Either that, or I'm not understanding what you need to do.

Do you need to find the values of x for which 4x^3 + 4x = 1 ?

**density** · September 23rd 2010, 11:10 AM

Originally Posted by Unknown008

The common factor is 4x. Can you try to factor it?

Either that, or I'm not understanding what you need to do.

Do you need to find the values of x for which 4x^3 + 4x = 1 ?

Yeah, I need to find x.
But I think that pulling 4x out is wrong.
$4x(x^2 + 1) = 1$
$4x=1 \quad x^2 + 1 = 1$ <------ isn't that illegal or something?

**e^(i*pi)** · September 23rd 2010, 11:20 AM

Indeed, you can't do that

You get $4x^3+4x-1 = 0$

There aren't any rational factors though

From wolfram: http://www.wolframalpha.com/input/?i=4x^3%2B4x-1+%3D+0

**Unknown008** · September 23rd 2010, 11:28 AM

Ok, no, that's not possible.

Bring all the terms on the same side.

$4x^3 + 4x= 1$

$4x^3 + 4x- 1=0$

However, this doesn't seem to have three real roots.

So, we'll go by iteration.

Set one x as subject of formula.

$4x = 1-4x^3$

$x = \dfrac{1-4x^3}{4}$

$x_{n+1} = \dfrac{1-4x_n\ ^3}{4}$

Lets take... x = 0.5

$x_1 = 0.125$

$x_2 =0.248$

$x_3 = 0.245$
.
.
.
$x \approx 0.236$

Thread: Solving a cubic equation.

LinkBack

Thread Tools

Display

Solving a cubic equation.

Similar Math Help Forum Discussions

Solving cubic equation

Solving a cubic equation.

Problem solving a cubic equation

Solving Cubic Equations

Solving cubic equation

Search Tags