If the roots of the equation $b{x}^2+cx+a=0$  are imaginary, then for all  real vaiues   of x, the expression 

$3b^2{x}^2+6bcx+2c^2$    is 

Given   $b{x}^2+cx+a=0$

Roots of Eq. (i) are imaginary i.e

  $C^2-4ab<0$            ……..( i ) 

$-4ab<-c^2\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{(}\mathit{ }ii\mathit{ }\mathit{)}\mathit{ }$

$\Rightarrow 3b^2{x}^2+6bcx+2c^2>\frac{-4ab<-c^2}{4A}=\frac{-\left(36b^2{c}^2-24b^2{c}^2\right)}{4\times 3b^2}=\frac{-12b^2{c}^2}{12b^2}$

 i.e  $3b^2{x}^2+6bcx+2c^2>-c^2,>-4ab$                                     [from Eq. (ii)]