Question

If $c<0$ and $a{x}^2+bx+c=0$ does not have any real roots then prove that $9a+3b+c<0$.

Answer 1

Prove the given expression

The given condition is $c<0$, $f\left(x\right)=a{x}^2+bx+c$

At $x=0,f\left(x\right)$is

$$\begin{gathered} f\left(0\right)=c \\ \therefore f\left(0\right)<0[\because c<0] \end{gathered}$$

Also, $f\left(x\right)$ does not have any real roots

$\therefore$ We can say for all $x\in \mathrm{ℝ},f\left(x\right)<0$

Let $x=3$

$$\begin{gathered} f\left(3\right)=9a+3b+c \\ \because f\left(x\right)<0\forall x\in \mathrm{ℝ} \\ \therefore f\left(3\right)<0 \\ \Rightarrow 9a+3b+c<0 \end{gathered}$$

Hence proved, $9a+3b+c<0$.