If $\alpha ,\beta$  are the 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 )

Since $\alpha ,\beta$  are roots of  Eq.( 1 ),  we get

  $\alpha +\beta =-p$  and $\alpha \beta =q.$

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

      $={\left[{\left(\alpha +\beta \right)}^2-2\alpha \beta \right]}^2-2q^2$

      $={\left[{\left(-p\right)}^2-2q\right]}^2-2q^2$

       $=p^4+4q^2-4p^{2}q-2q^2$

       $=p^4+2q^2-4p^{2}q$     …… ( 2 )

And  ${\alpha }^4{\beta }^4={\left(\alpha \beta \right)}^4=q^4$

As  ${\alpha }^4,{\beta }^4$  are roots of $x^2-rx+s=0,$  

we get ${\alpha }^4+{\beta }^4=r$  and  ${\alpha }^4{\beta }^4=s$

from Eq.( 2 ): $p^4+2q^2-4p^{2}q=r$       $\left(\because {\alpha }^4+{\beta }^4=r\right)$

And  now the discriminant of the equation $x^2-4qx+(2q^2-r)=0$  is

${(-4q)}^2-\mathrm{4.1.}(2q^2-r)$                

$=8q^2+4r=8q^2+4(p^4+2q^2-4p^{2}q)$

 $=16q^2-16p^{2}q+4p^4$

$=4{(P^2-2q)}^2\ge 0$

 $\Rightarrow$  the given equation has real roots.

Therefore, $\left(a\right)$ is the correct alternative.