If one root of the equation ax2+bx+c=0  is reciprocal of the one root of the equation a1x2+b1x+c1=0 , then

Let a be a root of the equation ax2+bx+c=0 Then , 1/α is a root of a1x2+b1x+c1=0 ⇒aα2+bα+c=0                          (i)and , a1α2+b1α+c1=0⇒c1α2+b1α+a1=0     (ii)From (i) and (ii), we have  α2ba1−b1c=αcc1−aa1=1ab1−c1b⇒a2=ba1−b1cab1−c1b,α=cc1−aa1ab1−c1b Now , α2=(α)2⇒ba1−b1cab1−c1b=cc1−aa12