Let x2+ax+b=0 has roots α and β x2−cx+d, roots are α4 and β4 α+β=−a,αβ=b and α4+β4=c,(αβ)4=d ⇒b4=d and α4+β4=c (α2+β2)2−2(αβ)2=c ⇒((α+β)2−2αβ)2−2(αβ)2=c ⇒(a2−2b)2−2b2=c⇒2b2+c=(a2−2b2) 2b2−c=4a2b−a2 =a2(4b−a2) Now for equation x2−4bx+2b2−c=0 D=(4b)2−4(1)(2b2−c) =16b2−8b2+4c =8b2+4c =4(2b2+c) =4(a2−2b)>0⇒ real roots Now f(0)=2b2−c =a2(4b−a2) <0 (since a2>4b) Roots are opposite in sign.