$|x^2-3x+2|=mx$ with 4 distinct real solutions, find range of $m$ I am working on my scholarship exam practice but not sure how to start here.

Find the range of $m$ such that the equation $|x^2-3x+2|=mx$ has $4$ distinct real solutions.

The answer provided is $0<m<3-2\sqrt{2}$.
Could you please give a solution or at least a hint to this question? 
 A: 
Let $m_0$ be a value when $y=m_0x$ touches $y=-x^2+3x-2$, then the answer is $m\in (0,m_0)$.
That is $$x^2+x(m-3)+2=0\implies (m-3)^2=8\implies m_0 = 3-\sqrt{8}$$

See why $\color{red}{m=3+\sqrt{8}}$ is not good:

A: Procedure (one of possibly many): On the region where we have
$$
|x^2-3x + 2| = x^2 - 3x + 2
$$
we are interested in the solutions to the equation
$$
x^2 - 3x + 2 = mx
$$
and on what region where we have
$$
|x^2 -3x + 2| = -x^2+3x-2\tag2
$$
we are interested in the solutions to the equation
$$
-x^2+3x-2 = mx
$$
Each of these two equations have at most two solutions. If we want four solutions, we need both of them to have two solutions. Which $m$ makes that happen?
A: Let $y=ax$, where $a>0$, be a tangent to the parabola $y=-x^2+3x-2$ on $(1,2)$.
Thus, we'll obtain the answer:
$$0<m<a.$$
Draw it!
A: Absolute values are not easy to deal with in practice, so start by noting that any solution to $|x^2-3x+2|=mx$ is either a solution to $$x^2-3x+2=mx$$ or a solution to $$x^2-3x+2=-mx\rm .$$
I am sure you will be able to find the non-negative values of $m$ for which the first equation has two real solutions, and also the non-negative values of $m$ for which the second equation has two solutions. The only thing you finally need to ensure, to satisfy the question, is that 2 solutions + 2 solutions = 4 solutions. And that (as you will see) is why the inequality is written as $0<m$, not allowing equality.
