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Polynomial equation

Let \Phi denote the golden ratio. Solve the equation

    \[x^3 +x^2-\Phi^5 x +\Phi^5=0\]

Solution

First of all we note that

    \[\left\{ \begin{array}{l} \Phi ^2 = \Phi + 1\\ \Phi ^3 = \Phi ^2 + \Phi = 2\Phi + 1 \end{array} \right. \Rightarrow \Phi ^5 = 2 \Phi ^2 + 3\Phi + 1 \Rightarrow \bold{\Phi ^5 = 5\Phi + 3}\]

We easily note that \Phi is one root of the equation, hence using Horner we get that

    \begin{align*} x^2 + \Phi ^2x - (2\Phi ^2 + \Phi ) = 0 &\Leftrightarrow x = \frac{ - \Phi ^2 \pm \sqrt {9\Phi ^2 + 6\Phi + 1}}{2} \\ &\Leftrightarrow x = \frac{ - (\Phi + 1) \pm (3\Phi + 1)}{2} \end{align*}

Hence \Phi is a double root and the other root is x=-2-\sqrt{5}.

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