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.
On
Let's start from $4a^2+3ab+2ac=a(4a+3b+2c)>0$.
Then we have
$8a^2+6ab+4ac>0$
$\implies 8a^2+6ab+b^2 > b^2-4ac$
$\implies \frac{8a^2+6ab+b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$ , since $a^2 > 0, a \in \mathcal{R} \wedge a \neq 0$
$\implies 2+\frac{3b}{2a}+\frac{b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$
$\implies (1+\frac{b}{2a})(2+\frac{b}{2a}) > \frac{b^2-4ac}{4a^2}$
$\implies (2+\frac{b}{2a})(2+\frac{b}{2a}) > (1+\frac{b}{2a})(2+\frac{b}{2a}) > \frac{b^2-4ac}{4a^2}$, since $2+\frac{b}{2a}>0$
$\implies (2+\frac{b}{2a})^2 > \frac{b^2-4ac}{4a^2}$
$\implies (2+\frac{b}{2a})^2 - \left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2 > 0$
$\implies \left(2+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right)\left(2+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right) > 0$
$\implies \left(-2-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right)\left(-2-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right) > 0$
$\implies \left(-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}-2\right)\left(-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}-2\right) > 0$
$\implies -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} > 2$ (in which case there is no root in between 1 and 2,since both roots > 2)
OR $\;\;-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} < 2$, in which case both the roots are less than 2. For the second case we need further analysis, can you take it from here?
Also, notice that $f(1)-(b+c)=a$ and $f(2)+(b+c)=4a+3b+2c$ have the same sign.
The roots are $-\frac{b}{2a}\pm\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}$, where the larger root $\geq$ 2, so that we may have at most one (the smaller) root in $(1,2)$, but it needs further analysis.
On
I took another approach with this after looking for a direct approach to working with the quadratic polynomial. We can define a quantity $ \ \Phi \ = \ \frac{4a + 3b + 2c}{a} \ \ , $ so that $ \ \Phi \ > \ 0 \ $ when both $ \ a \ $ and $ \ 4a + 3b + 2c \ $ have the same sign, and $ \ \Phi \ < \ 0 \ $ when they have opposite signs. For the polynomial $ \ ax^2 + bx + c \ $ with zeroes $ \ r \ $ and $ \ s \ \ , $ we have $$ \Phi \ \ = \ \ 4 \ + \ \frac{ 3b }{a} \ + \ \frac{2c}{a} \ \ = \ \ 4 \ - \ 3·(r + s) \ + \ 2·rs \ \ . $$
We can then plot $ \ \Phi \ $ in the " $ rs-$plane". The boundary between regions of positive [green] and negative [red] values of $ \ \Phi \ $ is the translated rectangular hyperbola $ \ 2rs \ - \ 3r \ - \ 3s \ + \ 4 \ = \ 0 \ \ $ centered on $ \ \left(\frac32 \ , \ \frac32 \right) \ $ and with asymptotes $ \ r \ = \ \frac32 \ \ , \ \ s \ = \ \frac32 \ \ $ (which has more than a little bearing on our problem). Since $ \ \Phi \ $ is independent of the ordering/designation of the two zeroes, we only need to consider the portion of the graph on or "above" the blue diagonal line.
We find that all pairs of zeroes $ \ 1 \ < \ r \ , \ s \ < \ 2 \ \ $ are contained within the red square, which lies entirely within the region $ \ \Phi \ < \ 0 \ \ . $ (The corners of the square tangent to the hyperbola correspond to the binomial-squares $ \ ( x - 1 )^2 \ $ and $ \ ( x - 2 )^2 \ \ . \ ) $ We can obtain this separately from $$ 1 \ < \ r \ , \ s \ < \ 2 \ \ \Rightarrow \ \ 2 \ < \ r + s \ < \ 4 \ \ , \ \ 1 \ < \ r·s \ < \ 4 $$ $$ \Rightarrow \ \ 6 \ < \ 4 \ + \ 2·r·s \ < \ 12 \ \ , \ \ -12 \ < \ -3·(r + s) \ < \ -6 \ \ \Rightarrow \ \ -6 \ < \ \Phi \ < \ 0 \ \ . $$ So we are able to say that when both zeroes of the quadratic polynomial lie in $ \ (1 \ , \ 2 ) \ \ , \ a \ $ and $ \ 4a + 3b + 2c \ $ have opposite signs . The contrapositive then holds (as observed in the analyses by Sandipan Dey and by mathlove ) : if $ \ \Phi \ > \ 0 \ \ , $ both zeroes cannot fall within that interval.
Otherwise, $ \ \Phi \ $ can tell us only a little more about the zeroes. The yellow bands on the graph indicate the situation for only one zero being found in $ \ (1 \ , \ 2 ) \ \ . $ Because the asymptotes bisect these bands, $ \ \Phi \ $ may be positive or negative for that case; further information is needed to resolve the sign of this quantity. Otherwise, we can say somewhat broadly that
• $ \ \Phi \ > \ 0 \ $ when $ \ r \ , \ s \ > \ 2 \ \ $ or $ \ \ r \ , \ s \ < \ 1 \ \ ; $
• $ \ \Phi \ < \ 0 \ $ when one zero is greater than $ \ 2 \ $ and the other is less than $ \ 1 \ \ . $
So $ \ \Phi \ $ can have either sign even when there are no zeroes within $ \ ( 1 \ , \ 2) \ \ . $ More precise investigation is needed if either zero is close to the boundary hyperbola.
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 $(1,2)$.