Finding the inequality satisfying the given conditions.

Question

(Multi-choice-multi-correct question):

If the roots of the equation $ax^2+bx+c=0$, $a\ne0$ are imaginary and $a+c<b$,($a,b,c$ are real numbers) then:

1. $a+4c< 2b$
2. $a+b+c<0$
3. $4a+c<2b$
4. $4a+c< 2b$ if $a<0$ and $4a+c>2b$ if $a>0$

My attempt:

$b^2-4ac<0 \implies a>0 \land c>0 $ or $a<0 \land c<0$

From condition 1 established above, (a>0,c>0) and given condition $a+c<b$

$4a+c+2b>0$ but this isn't the right answer. Where have I gone wrong? Any hints?

Answer 1

Roots are imaginary numbers $\implies b^2- 4ac < 0\implies b^2 < 4ac \le(a+c)^2\implies b^2 - (a+c)^2 < 0\implies (b-a-c)(b+a+c) < 0$

Since $b-a-c > 0$, it must be true that $a+b+c < 0$.

Thus $2)$ is true.