inequality – Proving on the equation $(x^2+mx+n)(x^2+px+q)=0$

Find all real numbers k such that if $a,b,c,d in mathbb R$ and $a>b>c>d geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that the following equation has 4 distinct real solutions :

$(x^2+mx+n)(x^2+px+q)=0$

Here all i did :

$(x^2+mx+n)(x^2+px+q)=0$

$Leftrightarrow x^2+mx+n=0$ or $x^2+px+q=0$

so I think the four real solutions, if any, of the equation can only be :

  • $ x= frac{sqrt{m^2-4n} -m}{2} $

  • $ x= frac{-sqrt{m^2-4n} -m}{2} $

  • $ x= frac{sqrt{p^2-4q} -p}{2} $

  • $ x= frac{-sqrt{p^2-4q} -p}{2} $

So I think to solve the problem we just need to find all real numbers k such that if $a,b,c,d in mathbb R$ and $a>b>c>d geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that
$p^2 geq 4q$ and $ m^2 geq 4n$

We can prove that for $k >= 4 $ it is absolutely true. So are there any other satisfying $k $ values ? I’m not entirely sure. Hope to get help from everyone. Thanks very much !