- Find the range of $m$ such that the equation $|x ^ 2 - 3x + 2| = mx$ has $4$ distinct real solutions $\alpha$, $\beta$, $\gamma$, and $\delta$.
- Express the value $s(m) = 1/\alpha ^ 2 + 1/\beta ^ 2 + 1/\gamma ^ 2 + 1/\delta ^ 2$ in terms of $m$.
- When $m$ varies as in $(1.)$, find the range of $s(m)$.
This is a question from the Japanese government MEXT scholarship $2013$ paper.
What I did so far:
The range I found for $m$ is $0 < m < 3 - 2\sqrt{2}.$ I found it by finding the maximum slope of the tangent to the parabola $-(x^2 - 3x + 2)$ as it would be the limiting value for $m.$
I could solve the first part but cannot understand how to approach the $2$nd and $3$rd sub-question. Please help me with some hints on how to approach the questions. Thanks in advance!
The solutions are $$\alpha=\frac{3+m+\sqrt{D_1}}{2},\; \beta=\frac{3+m-\sqrt{D_1}}{2},\;\gamma=\frac{3-m+\sqrt{D_2}}{2},\;\delta=\frac{3-m-\sqrt{D_2}}{2},$$ where $D_1=m^2+6m+1>0$ and $D_2=m^2-6m+1>0.$
HINT for question $2.$
We obtain by straightforward computation $$\alpha\beta=2\quad \text{and}\quad \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3+m}{2}.$$ Then $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^2-1$$
Analogous approach gives $\left(\frac{1}{\gamma^2}+\frac{1}{\delta^2}\right)$ in terms of $m.$