By using the substitution $p=x+\frac{1}{x}$, show that the equation $$2x^4+x^3-6x^2+x+2=0$$ reduces to $2p^2+p-10=0$.

I can't think of anything that produces a useful result, I tried writing p as $p=\frac{x^2+1}{x}$ and finding areas to substitute but have come with no progress. Could someone offer a slight hint on how to proceed?

Thanks

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Like this,

as the equation is Reciprocal Equation of the First type with $x\ne0,$

divide either sides by $\displaystyle x^{\frac42}=x^2$ to reduce the degree of the eqaution by half

$$2x^2+x-6+\frac1x+\frac2{x^2}=0$$

$$\implies 2\left(x^2+\frac1{x^2}\right)+\left(x+\frac1x\right)-6=0$$

$$\implies 2\{\left(x+\frac1x\right)^2-2\}+\left(x+\frac1x\right)-6=0$$

Reference : Reciprocal Equation is explained here:

Chapter XI of Higher Algebra,Barnard & Child and

Article $568−570$ of Higher algebra, Hall & Knight

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how do I know how to write $x^2+\frac{1}{x^2}$ as $(x+\frac{1}{x})^2-2$? –  salman Aug 30 '13 at 18:03
@user90771, as you have $p=x+\frac1x,$ and we know $(x+\frac1x)^2=x^2+\frac1{x^2}+2$ –  lab bhattacharjee Aug 30 '13 at 18:07
also jee, how do I know that I must divide by $x^2$? thanks –  salman Aug 30 '13 at 18:08
@user90771, please find the edited answer, hope this should explain the idea –  lab bhattacharjee Aug 30 '13 at 18:11
this is an amazing resource thanks bhai –  salman Aug 30 '13 at 18:24

\begin{align} 2x^4+x^3-6x^2+x+2 & = x^2\Big(2x^2 + x - 6 + \frac1x + \frac{2}{x^2}\Big) \\[12pt] & = x^2 \Big(2\left(x+\frac1x\right)^2 + \left(x+\frac1x\right) - 10 \Big) \\[12pt] & = x^2(2p^2 +p-10). \end{align} This equals $0$ only if $x=0$ or the second factor equals $0$. But clearly $x=0$ is not one of the solutions.

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$$p=x+x^{-1}$$ $$p^2=x^2+2+x^{-2}$$ $$2x^4+x^3-6x^2+x+2=0$$ $$2x^2+x-6+x^{-1}+2x^{-2}=0$$ $$2(x^2+2+x^{-2})+(x+x^{-1})-10=0$$ $$2p^2+p-10=0$$

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what indicates that I must divide by $x^2$? –  salman Aug 30 '13 at 18:06
@user90771 It's the standard usual method which works well. –  namehere Aug 30 '13 at 18:07
@user90771 You divide by x square because it just happens to work. In general quartic equations are not subject to substitutions as easy as this one. –  namehere Aug 30 '13 at 18:08

Hint: Try working backwards. Start with: $$2\left(x + \dfrac{1}{x}\right)^2 + \left(x + \dfrac{1}{x}\right) - 10 = 0$$ then try to obtain the original equation by expanding then clearing the denominators.

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