It is given that, $\alpha , \beta$ are the roots of $x^2+px+q=0$
$\alpha +\beta =-p$ and $\alpha \beta =q$
Since, ${\alpha }^4, {\beta }^4$ are roots of $x^2-rx+s=0$

Therefore, ${\alpha }^4+{\beta }^4=r$ and ${\alpha }^4{\beta }^4=s$
Now, $x^2-4qx+2q^2-r=0$

$D=(4q{)}^2-4\left(2q^2-r\right)$

$=16q^2-8q^2+4r$

$=8q^2+4r$

Here, $r={\alpha }^4+{\beta }^4\ge 0$ and $8q^2\ge 0.$
Thus, $D\ge 0$

Therefore, the equation has two real roots.