If both roots of equation $ax^2+x+c-a=0$ are imaginary and $c>-1$ then
(A) $3a > 2 + 4c$
(B) $3a < 2 + 4c$
(C) $c < a$
(D) none of these
My approach is as follow
${D^2} = 1 - 4a\left( {c - a} \right) < 0$ for imaginary roots condition
Hence $4a(c-a)>1$
$4{a^2} - 4ac + {c^2} + 1 - {c^2} < 0 \Rightarrow {\left( {c - 2a} \right)^2} + 1 - {c^2} < 0$
How do I approach from here