Rearranging help?

**Drummerdude93** · September 30th 2010, 01:30 PM

Alright, this has stumped me.

How can I rearrange:

2x^3 - x - 4 = 0

to be:

x = ROOT(2/x + 1/2) ?

**Archie Meade** · September 30th 2010, 01:54 PM

Originally Posted by Drummerdude93

Alright, this has stumped me.

How can I rearrange:

2x^3 - x - 4 = 0

to be:

x = ROOT(2/x + 1/2) ?

Hi Drummerdude,

$2x^3-x-4=0$

$2x^3-x=4$

Factor...

$x\left(2x^2-1\right)=4$

$2x^2-1=\frac{4}{x}$

$2x^2=\frac{4}{x}+1$

$x^2=\frac{2}{x}+\frac{1}{2}$

Finally, take the square root.

**skeeter** · September 30th 2010, 01:59 PM

Originally Posted by Drummerdude93

Alright, this has stumped me.

How can I rearrange:

2x^3 - x - 4 = 0

to be:

x = ROOT(2/x + 1/2) ?

note that the transformation is incorrect ... it should be $x = \pm \sqrt{\frac{2}{x} + \frac{1}{2}} \, ; x \ne 0$

what problem are you solving that led you to ask about this?

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