If $\alpha and\beta$ are the real roots of $x^2+px+q=0$  and ${\alpha }^4,{\beta }^4$ are the roots of $x^2-rx+s=0,$  then the equation $x^2-4qx+2q^2-r=0$  has always

The given equation is $x^2+px+q=0$  …. ( 1 )

$\alpha and\beta$ are the real roots of ( 1 )

Then sum of the roots : $\alpha +\beta =-p,$

Product of the roots :$\alpha \beta =q$

And

The given another equation is $x^2-rx+s=0$  …. ( 2 )

${\alpha }^4,{\beta }^4$ are the roots of ( 2 )

then sum of the roots :${\alpha }^4+{\beta }^4=r,$

product of the roots : ${\alpha }^4{\beta }^4=s$

$\therefore$  $r>0.$  For the equation $x^2-4qx+2q^2-r=0,$

Here $a=1,b=-4q,c=2q^2-r$

Dicriminant $D=16q^2-4\left(2q^2-r\right)=8q^2+4r>0$        

      $=8{\alpha }^2{\beta }^2+4\left({\alpha }^4+{\beta }^4\right)=4{\left({\alpha }^2+{\beta }^2\right)}^2$

$\Rightarrow$    equation has two real roots.