If the roots of the equation  b x2+c x+a=0  be imaginary, then for all real values of  x, the expression  3b2x2+6bc x+2c2  is

Given equation is c2−4ab<0 ….. ( 1 )Given that the roots of  Eq. ( 1 )  are  Imaginary (non-real),therefore, (∵ b2−4ac<0)        ⇒    c2<4ab3b2x2+6bcx+2c2.and the given expression is =  3b2[x2+2cbx]+2c2  =  3b2[x2+2cbx+c2b2−c2b2]+2c2        add and subtractc2b2          =  3b2  (x+cb)2+2c2−3b2(c2b2)    =  3b2(x+cb)2−c2     ≥ −c2>−4ab          (∵  c2<4ab )