If $ a(12a + 5b +2c) > 0 $ , then prove that the real roots of equation $ ax^2+ bx +c =0$ are less than 2

Question

I know that to prove both roots less than 2 ,i have to prove $ 4a+ 2b +c > 0$ and $ -b/2a < 2 $ . Here i have no idea how to proceed.

Answer 1

The original statement is not true. Take $\mathbb p(x)=2x^2-5x+1$. Clearly $a(12a+5b+2c)=2>0$ but one of the roots ($\approx 2.281$) is greater than $2$.

The following proves the converse:

Note that if both roots are less than $2$, then:

$(1)$ Both roots must be on the same side of $2$ on the x- axis.

$(2)$ $2$ must be greater than the mid-point of the two roots.

Both the above conditions ensure that both roots are less than $2$. Draw possible graphs of $\mathbb p(x)=ax^2+bx+c$ to get a feel of this.

$(1)$ is ensured by $a\times \mathbb p(2)>0$.

$(2)$ is ensured by $2 > \dfrac 12\left (\dfrac{-b+ \sqrt{b^2-4ac}}{2a}+\dfrac{-b- \sqrt{b^2-4ac}}{2a}\right )$.

We therefore get $$a(4a+2b+c)>0\iff a(8a+4b+2c)>0$$ and the second condition gives us $$4>-\dfrac ba\iff 4a^2>-ab\iff a(4a+b)>0$$ Adding the two conditions obtained, we get $$a(12a+5b+2c)>0$$