Question

Given $a , b , c$ are three distinct real numbers satisfying the inequality $a -2 b +4 c >0$ and the equation $ax ^2+ bx + c =0$ has no real roots. Then the possible value of $\frac{4 a +2 b + c }{ a +3 b +9 c }$ is/are

• A. 2
• B. -1
• C. 3
• D. $\sqrt{2}$

Answer 1

Answer: (A), (C)

Let $f(x)=a^2+b x+c$
Since equation $ax ^2+ bx + c =0$ has no real roots, so
$f ( x )>0 \forall x \in R$ or $f ( x )<0 \forall x \in R$
But given $a-2 b+4 c>0 \Rightarrow f\left(\frac{-1}{2}\right)>0 \Rightarrow f(x)>0 \forall x \in R$.
$\therefore f (2)=4 a +2 b + c >0$ and $f \left(\frac{1}{3}\right)=\frac{ a +3 b +9 c }{9}>0$
So, $ \frac{f(2)}{9 f\left(\frac{1}{3}\right)}=\frac{4 a+2 b+c}{9 c+3 b+a}>0$
Hence $\frac{4 a+2 b+c}{9 c+3 b+a}>0$.