If the two equations a1x2+b1x+c1=0  and a2x2+b2x+c2=0  have a common root, then the value of (a1b2−a2b1) (b1c2−b2c1)=

The given equations are a1x2+b1x+c1=0 And a2x2+b2x+c2=0 Given that α  is a common root of  EquationsThen the equations are a1α2+b1α+c1=0 ….. ( 1 )α2α2+b2α+c2=0 ….. ( 2 )Solve  ( 1 )&( 2 )⇒    α2b1c2−c1b2  =  αc1a2−a1c2=  1a1b2−a2b1 ⇒    α2α  =  b1c2−c1b2c1a2−a1c2   and α1=  c1a2−a1c2a1b2−a2b1 ⇒    b1c2−c1b2c1a2−a1c2  =  c1a2−a1c2a1b2−a2b1. ⇒(a1b2−a2b1) (b1c2−b2c1)= (c1a2−a1c2)(c1a2−a1c2) ∴(a1b2−a2b1) (b1c2−b2c1)= (a1c2−c1a2)2