<s>[INST] Number of roots of the equation $ax^2+ bx + c = 0$ in $(1,2)$ 
Let $a, b, c \in R, a \ne 0$ such that $a$ and $4a + 3b + 2c$ have the same sign. Then the
number of roots of the equation $ax^2+ bx + c = 0$ that lie(s) in $(1,2)$ is(are)?

I began by writing that $4a^2+3ab+2ac>0$. I tried finding the sign of $f(1)\cdot f(2)$ to use this condition and Bolzano's theorem. Unfortunately, I couldn't find a way to proceed.
 [/INST] One can say that if the number of the roots lying in $(1,2)$ is $2$, then the axis of symmetry of the parabola $y=ax^2+bx+c$ lies in $(1,2)$.
Therefore, one can say that if the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$, then the number of the roots lying in $(1,2)$ is not $2$.
One has
$$0\lt \frac{2a(4a+3b+2c)}{4a^2}\leqslant \frac{8a^2+6ab+b^2}{4a^2}=\bigg(1-\bigg(-\frac{b}{2a}\bigg)\bigg)\bigg(2-\bigg(-\frac{b}{2a}\bigg)\bigg)$$
i.e.
$$\bigg(1-\bigg(-\frac{b}{2a}\bigg)\bigg)\bigg(2-\bigg(-\frac{b}{2a}\bigg)\bigg)\gt 0$$
which means that the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$.
Therefore, one can say that the number of the roots lying in $(1,2)$ is not $2$.
If $a=1,b=1$ and $c=-1$ satisfying $a(4a+3b+2c)\gt 0$, then the number of roots lying in $(1,2)$ is $0$.
If $a=1,b=1$ and $c=-3$ satisfying $a(4a+3b+2c)\gt 0$, then the number of roots lying in $(1,2)$ is $1$.

Added :
If $a(4a+3b+2c)\gt 0$ and $b^2-4ac=0$, then the number of the roots lying in $(1,2)$ is $0$.
Under $a(4a+3b+2c)\gt 0$ and $b^2-4ac\gt 0$, the number of the roots lying in $(1,2)$ is
$$\begin{cases}0&\text{if $\ (a+b+c)(4a+2b+c)\geqslant 0$}
\\\\ 1&\text{otherwise}\end{cases}$$
This can be seen by noting that the axis of symmetry of the parabola $y=ax^2+bx+c$ does not lie in $(1,2)$ and that $y=ax^2+bx+c$ is monotone on $(1,2)$.
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