Alright, this has stumped me. How can I rearrange: 2x^3 - x - 4 = 0 to be: x = ROOT(2/x + 1/2) ?
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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, $\displaystyle 2x^3-x-4=0$ $\displaystyle 2x^3-x=4$ Factor... $\displaystyle x\left(2x^2-1\right)=4$ $\displaystyle 2x^2-1=\frac{4}{x}$ $\displaystyle 2x^2=\frac{4}{x}+1$ $\displaystyle x^2=\frac{2}{x}+\frac{1}{2}$ Finally, take the square root.
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 $\displaystyle x = \pm \sqrt{\frac{2}{x} + \frac{1}{2}} \, ; x \ne 0$ what problem are you solving that led you to ask about this?
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