Let me ask it differently:
1st example:
I have the polynomial $$p(x)=x^4 - \sqrt{\frac{\epsilon}{16} + 8} \, x^3 + r\,x^2 + r\sqrt{\frac{\epsilon}{16}+8} \, x - 2r^3$$ which has 4 roots, three of which are $>0$. The graph of the three positive roots as a function of $r>0$ for fixed $\epsilon$ has 2 disconnected parts. In the case $\epsilon=0$ it becomes degenerate i.e. the 2 disconnected parts meet at $r=1$. For $\epsilon>0$ there is one larger branch and the two lower branches build together 1 connceted graph. Likewise for $\epsilon<0$ the 2 larger branches build together 1 connected graph, while the lowest branch stays separated. Therefore the point $\epsilon=0,r=1$ is a bifurcation point.
Now my question is, if it is coincidence that this bifurcation point $\epsilon=0,r=1$ is precisely the point where not only the discriminant $D(\epsilon,r)$ of $p(x)$ vanishes, but also the derivative with respect to $r$ $$D(\epsilon,1)={\frac {\epsilon\, \left( \epsilon-16 \right) ^{2}}{1024}}\\ \partial_r D(\epsilon,1) = {\frac {\epsilon\, \left( 3\,{\epsilon}^{2}-1184\,\epsilon-18688 \right) }{1024}} \, .$$
2nd example: The polynomial is $$p(x)=x^3+2rx^2+rx+\epsilon$$ and its discriminant + derivative is $$D(\epsilon,r)=-32\,\epsilon\,{r}^{3}+4\,{r}^{4}+36\,\epsilon\,{r}^{2}-4\,{r}^{3}-27\,{\epsilon}^{2} \\ \partial_r D(\epsilon,r)=-96\,\epsilon\,{r}^{2}+16\,{r}^{3}+72\,\epsilon\,r-12\,{r}^{2} \, .$$ The graph of the three solutions degenerates (connected branches change) at $\epsilon=0, r=0$ and as it can be seen the discriminant and its derivative vanish at this point.
Coincidence?
Your $p_3(x;\alpha)$ could vary with respect to $\alpha$ in such a way that two roots coincide but not the third, e.g. $p_3(x;\alpha)=x^3-2\alpha x^2+\alpha^2$. Then $D=0$ identically but of course the only time when three roots coalesce is at $\alpha=0$ which is invisible to $D$.
Or you could have $p_3(x;\alpha)=(x-\alpha)^2(x-\alpha-1)=x^3-(3\alpha+1)x^2+(3\alpha^2+2\alpha)x-\alpha^2(\alpha+1)$, then again $D$ is identically 0 but you never get all three roots coalesce.
You need to go back to, e.g., $p,p',p''$ all share a root.