# Math Help - Solving a cubic equation.

1. ## Solving a cubic equation.

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

and am trying to solve

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

2. 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 ?

3. 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?

4. 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

5. 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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$x \approx 0.236$