# How to find the equation with roots $\alpha/\beta$ and $\beta/\alpha$, given that $\alpha \ne \beta$, $\alpha^2 = 5\alpha -3$, $\beta^2 = 5\beta -3$

I have $\alpha \ne \beta$, $\alpha^2 = 5\alpha -3$, $\beta^2 = 5\beta -3$.

I need to find out an equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.

How to find out?

$\alpha+\beta=5$ and $\alpha\beta=3$.
Thus, $$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{25-6}{3}=\frac{19}{3}$$ and $$\frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}=1,$$ which gives the answer: $$3x^2-19x+3=0.$$
You can find original equation $$\alpha^2 = 5\alpha -3\\\beta^2 = 5\beta -3 \\\to x^2=5x-3 \\\to x^2-5x+3=0\to s=\frac{-b}{a}=+5 ,p=\frac{c}{a}=+3$$ so you need an equation with $$\alpha'=\frac{\alpha}{\beta},\beta'=\frac{\beta}{\alpha}$$so you need to find $$s',p'$$to put into $$x^2-s'x+p'=0$$ $$s'=\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{s^2-2p}{p}=\frac{25-6}{3}=\frac{19}{3}\\p'=\frac{\alpha}{\beta}\cdot\frac{\beta}{\alpha}=1$$ so $$x^2-\frac{19}{3}x+1=0 \to (\times 3)\\3x^2-19x+3=0$$