Given that roots of the equation bx2+cx+a=0 are imaginary ∴c2−4ab<0 Let y=3b2x2+6bcx+2c2 ⇒3b2x2+6bcx+2c2−y=0 As x is real, D≥0 ⇒36b2c2−12b2(2c2−y)≥0 ⇒12b2(3c2−2c2+y)≥0[∵b2≥0] ⇒c2+y≥0⇒y≥−c2 But from eqn. (i), c2<4ab or −c2>−4ab ∴ we get y≥−c2>−4ab ⇒y>−4ab