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**NonCommAlg** so you have: $\displaystyle \begin{cases} \alpha + \beta=-p \\ \alpha \beta = q \\ k\alpha + \frac{\beta}{k}=-p' \end{cases}.$ fro the last equation you get $\displaystyle \alpha k^2 + \beta=-p'k,$ which with the first equation give us: $\displaystyle \alpha=\frac{p-p'k}{k^2-1}, \ \beta=\frac{(p'-pk)k}{k^2 - 1}.$ [here i'm assuming that $\displaystyle k \neq \pm 1.$]

putting what we got for $\displaystyle \alpha, \beta$ in the second equation, i.e. $\displaystyle \alpha \beta = q,$ gives us: $\displaystyle (p-p'k)(p'-pk)k=q(k^2-1)^2.$ thus: $\displaystyle qk^4 -pp'k^3 + (p^2 + p'^2 - 2q)k^2 - pp'k + q=0$ and the result follows.