If α and β are the roots of the equation x2+ax+b=0, and α4 and β4 are the roots of x2-px+q=0, the roots x2−4bx+2b2−p=0 are always

We have,     α+β=−a, αβ=b, α4+β4=p and α4β4=q.Now,      α4+β4=α2+β22−2α2β2⇒   α4+β4=(α+β)2−2αβ2−2(αβ)2⇒   α4+β4=a2−2b2−2b2         [∵α+β=−a,αβ=b]⇒   α4+β4=a22−4a2b+2b2⇒   p=a22−4ba2+2b2             ∵α4+β4=p      ⇒ a22−4ba2+2b2−p=0          …(i)      ⇒ a2 is a root of the equation x2−4bx+2b2−p=0Let D be the discriminant of x2−4bx+2b2−p=0. Then,    D=16b2−42b2−p=16b2+4a22−4ba2 [Using (i)]⇒D=4a22−4ba2+4b2=4a2−2b2≥0Hence, the roots of the given equation are real and are of opposite signs.