# Solving a cubic equation.

• Sep 23rd 2010, 11:59 AM
density
Solving a cubic equation.
I have $f(x)= x^4 + 2x^2$

and am trying to solve

$f'(x)= 1$, that is $4x^3 + 4x= 1$
• Sep 23rd 2010, 12:03 PM
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 ?
• Sep 23rd 2010, 12:10 PM
density
Quote:

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?
• Sep 23rd 2010, 12:20 PM
e^(i*pi)
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
• Sep 23rd 2010, 12:28 PM
Unknown008
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

$x_1 = 0.125$

$x_2 =0.248$

$x_3 = 0.245$
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$x \approx 0.236$