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- #1

- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,802

- Thread starter
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- #2

- Feb 14, 2012

- 3,802

$\begin{align*}P(Q(x))&=\left(x^2+\dfrac{x}{2}\right)^2+\dfrac{1}{2}\left(x^2+\dfrac{x}{2}\right)-\dfrac{1}{2}\\&=\dfrac{1}{4}(4x^4+4x^3+3x^2+x-2)\end{align*}$

Clearly $P(Q(-1))=0$ and any other rational root will be $\pm\dfrac{1}{4},\,\pm \dfrac{1}{2}$. Of these, $\dfrac{1}{2}$ is a root. The remaining factor is $h(x)=x^2+\dfrac{x}{2}+1$ and this has no real roots.