Since, a+b=−p,ab=1....(i) and c+d=−q,cd=1 Now (a−c)(b−c) and (a+d)(b+d) are the roots of x2+ax+β=0 (a−c)(bc)(a+d)(b+d)=β ⇒(ab−ac−bc+c2)(ab+ad+bd+d2)=β ⇒{1−c(a+b)+c2}{1+d(a+b)+d2}=β ⇒(1+pc+c2)(1−pd+d2)=β ⇒1−pd+d2+pc−p2cd+pcd2+c2−pc2d+c2d2=β ⇒1−pd+d2+pc−p2+pd+c2−pc+1=β[∵cd=1] ⇒2+d2+c2−p2=β ⇒2cd+c2+d2−p2=β (c+d)2−p2=β q2−p2=β[∵(c+d)=−q]