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The question:

Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.

This problem is seems to be related to Vieta's Theorem, but so far I have not used it. This is my working out:

\begin{align} \frac {\alpha}{\beta} + \frac {\beta}{\alpha} & = \frac {\alpha^2}{\alpha \beta} + \frac {\beta^2}{\alpha \beta} \\ & = \frac {\alpha^2+\beta^2}{\alpha \beta} \end{align}

Vieta's Theorem:

Let $p(x)=ax^2+bx+c$ be a quadratic polynomial with zeros $\alpha$,$~\beta$. Then $$\frac {-b}{a}=\alpha + \beta \\ \frac{c}{a} = \alpha \cdot \beta $$

Well, it is clear that the question wants us to find $\alpha^2 + \beta^2$ and $\alpha \beta$.

\begin{align} \alpha^2+\beta^2 & = \alpha^2+2\alpha \beta + \beta^2 -2\alpha\beta \\ & = (\alpha+\beta)^2-2(\alpha\beta) \end{align}


We have:

\begin{align} \alpha^2 + 3\alpha + 1 & = \beta^2 + 3\beta + 1 \\ 0 & =\alpha^2 - \beta^2+3\alpha-3\beta \\ & = (\alpha+\beta)(\alpha-\beta)+3(\alpha-\beta) \\ & = (\alpha-\beta)(\alpha+\beta+3) \\ \therefore~ \alpha + \beta & = -3\tag{reject $\alpha=\beta$} \end{align}

This is where I am stuck because I am unable to find $\alpha\beta$ by algebraic manipulation. I have thought about trying to find $\alpha$ and $\beta$ through the quadratic formula, but it seems quite tedious, so it is a last resort. Is there a method to finish this question off?

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Since $\alpha $ and $\beta $ are roots of $x^2+3x+1$ we know that $\alpha +\beta = -3$ and $\alpha \beta = 1$ Indeed $$ x^2+\color{red}{3}x+\color{blue}{1} =(x-\alpha)(x-\beta) = x^2 -\color{red}{(\alpha +\beta)}x+\color{blue}{\alpha \beta}$$

Hence,

$$\frac {\alpha}{\beta} + \frac {\beta}{\alpha} = \frac {\alpha^2+\beta^2}{\alpha \beta} = \frac {(\alpha +\beta)^2 -2\alpha \beta}{\alpha \beta} = \frac{(-3)^2-2}{1}=\color{blue}{7}$$

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  • $\begingroup$ Oh, of course, since $f(\alpha)=0$ and $f(\beta)=0$. I overlooked that detail. Thanks! $\endgroup$ – Landuros Dec 1 '17 at 13:27
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You have already written down $\frac{c}{a} = \alpha \cdot \beta$, so there you have it.

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$$\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$$

Implies that $\alpha$ and $\beta$ are the roots of $x^2 + 3x + 1$, thus:

$$x^2 + 3x + 1 = (x - \alpha)(x - \beta)$$

Expand $(x - \alpha)(x - \beta)$ to find $-\alpha - \beta = 3$ and $\alpha\beta = 1$.

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