Question
The roots of $$a{x^2} + bx + c = 0,$$ where $$a \ne 0$$ and co-efficients are real, are non-real complex and $$a + c < b.$$ Then
A.
$$4a + c > 2b$$
B.
$$4a + c < 2b$$
C.
$$4a + c = 2b$$
D.
None of these
Answer :
$$4a + c < 2b$$
Solution :
$$\eqalign{
& {b^2} - 4ac < 0\,\,{\text{and }}a + c < b. \cr
& {b^2} - 4ac < 0 \cr} $$
⇒ $$a, c$$ must be of the same sign.
If $$a, c$$ are both positive, $${\left( {a + c} \right)^2} < {b^2}.$$
$$\eqalign{
& \therefore \,\,{b^2} - 4ac < 0 \cr
& \Rightarrow \,\,{\left( {a + c} \right)^2} - 4ac < 0 \cr
& \Rightarrow \,\,{\left( {a - c} \right)^2} < 0\,\,{\text{which is absurd}}{\text{.}} \cr} $$
So $$a, c$$ will both be negative. For example, $$ - {x^2} + 4x - 5 = 0.$$
It satisfies, $$D < 0$$ and $$a + c < b.$$ Also $$4a + c = - 4 - 5 = - 9 < 8 = 2b.$$
Clearly, when $$a, c$$ are negative, $$b$$ must be positive. Otherwise, the
equation becomes one where co-efficients are all positive.