Question

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

• A. 2
• B. -1
• C. 3
• D. $\sqrt{2}$

Answer 1

Answer: (A), (C)

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