...(i) and α4 and β4 are the roots x2−rx+3=0 α4+β4=r α4β4=s...(ii) Now , 2x2−4p2x+4p−2r=0 D=B2−4AC ∴D=16p4−4[2(p2−2r)] =16p4−8p4+16r =8(p4+2r) =8[4α2β2+2(α4+β4)] =16[(α+β)2−2αβ]2 =16(p2−p2)=0 Hence, the roots of the equation 2x2−4p2x+4p−2r=0 are equal