If α  and  β are the real roots of x2+px+q=0  and α4, β4 are the roots of x2−rx+s=0,  then the equation x2−4qx+2q2−r=0  has always

The given equation is x2+px+q=0  …. ( 1 ) α  and  β are the real roots of ( 1 )Then sum of the roots : α+β=−p,  Product of the roots : αβ=q AndThe given another equation is x2−rx+s=0  …. ( 2 )α4, β4 are the roots of ( 2 )then sum of the roots : α4+β4=r,  product of the roots : α4β4=s ∴  r>0.  For the equation x2−4qx+2q2−r=0, Here a=1,b=−4q,c=2q2−r Dicriminant D=16q2−4(2q2−r)=8q2+4r>0                  =8α2β2+4(α4+β4)=4(α2+β2)2 ⇒    equation has two real roots.