Find x

**jkami** · February 28th 2009, 05:27 PM

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

Ok, here is how I tried to solve and where I stucked at

$4x^3-4x+1=0$
$4x(x^2-1)+1=0$
$4x(x-1)(x+1)=-1$

I don't know what to do next

**Krizalid** · February 28th 2009, 05:52 PM

Alas, that won't work, but rewrite $-4x$ as $-2x-2x$ and factor by grouping.

**jkami** · February 28th 2009, 05:57 PM

ok, so

$4x^3-4x+1=0$
$4x^3-2x-2x+1=0$
$2x(2x^2-1)-(2x-1)=0$

hmmm...

**Krizalid** · February 28th 2009, 06:00 PM

Ohhh, sorry, I thought it was a $x^2.$

I'll think in other method.

**jkami** · February 28th 2009, 06:39 PM

I tried to graph it out, x should have 3 values...

**mr fantastic** · February 28th 2009, 08:52 PM

Originally Posted by jkami

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

Ok, here is how I tried to solve and where I stucked at

$4x^3-4x+1=0$
$4x(x^2-1)+1=0$
$4x(x-1)(x+1)=-1$

I don't know what to do next

Are you expected to get exact solutions by hand? In which case you have a difficult job ahead because I don't think there are rational roots. You will need to follow the process explained here: The "Cubic Formula"

Or are decimal approximations sufficient? In which case, use technology to get solutions to the required accuracy.

Posting the whole question, including the context it's taken from, will shed light on this if you cannot answer these questions.

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