Roots of quadratic with $a$ and $8a+3b+c$ of the same sign cannot lie in (2,4)

Question

I have the second degree equation $ax^2+bx+c=0$ with $a$ non-zero such that $a$ and $8a+3b+c$ have the same sign and I need to prove that both the roots cannot lie in $(2,4)$.

My try:

I assume two roots are in $(2,4)$

$2 < \frac{-b+\sqrt{b^2-4ac}}{2a} < 4$

$2 < \frac{-b-\sqrt{b^2-4ac}}{2a}< 4$

$2+\frac{b}{2a}<\frac{\sqrt{b^2-4ac}}{2a} < 4+\frac{b}{2a}$

and

$2+\frac{b}{2a}<-\frac{\sqrt{b^2-4ac}}{2a} < 4+\frac{b}{2a}$

$\implies 2+\frac{b}{2a}<|\frac{\sqrt{b^2-4ac}}{2a}|<4+\frac{b}{2a}$

$\implies (2+\frac{b}{2a})^2<\frac{b^2-4ac}{4a^2}<(4+\frac{b}{2a})^2$

$\implies 4+4\frac{b}{a}+\frac{b^2}{4a}<\frac{b^2-4ac}{4a^2}<16+4\frac{b}{a}+\frac{b^2}{2a^2}$

$\implies 4+4\frac{b}{a}<\frac{-c}{a}<16+4\frac{b}{a}$

What to do next? How to apply condition?

Answer 1

Hint: Let's say the roots are $x_1$ and $x_2$. If the roots lie in $(2,4)$ simultaneously, then:

$$(x_1-2)(x_2-4)+(x_1-4)(x_2-2)<0$$

and then Vieta's. Can you end it from here?