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

I think that $(x^2-3x+2) = mx$ or $-(x^2-3x+2) = mx$
For $-(x^2-3x+2) = mx$, $D > 0$,
$m = 3 \pm \sqrt{8}$
For $(x^2-3x+2) = mx$, $D > 0$,
$m = -3 \pm \sqrt{8}$
Now. What do you think it's the range?
Note that $ |x^2-3x+2 | = mx$ which means $mx > 0$ isn't it?

Solving $|x^2-3x+2|=mx$ is same as finding the intersection of the graph of $y=|x^2-3x+2|$ and the lines $y=mx$ passing through the origin. Here is a figure to better illustrate the idea.

By changing the value of $m$ we are changing the slope of the line passing through the origin. Unless this line is tangent or  passes through the middle cusp in the graph there cannot be $4$ solutions. This means the first instance the line will intersect the graph at $4$ points (tangent point is counted as double point) is when this line is tangent to the graph. 
Ques: what is the point of tangency?
Let $(h,k)$ be the point on (the middle cusp) the curve such that the line $y=mx$ is tangent. Then using the fact that slope of the line $\frac{k}{h}$ should be the same as the slope of the curve (in particular the middle cusp which is given by $y=-x^2+3x-2$), we get
$$m=\frac{k}{h}=\frac{dy}{dx}\Big|_{(h,k)}=-2h+3.$$
But $k=-h^2+3h-2$, so this gives $h=\pm \sqrt{2}$. Note that the negative value is not possible. So $\color{red}{h=\sqrt{2}}$.
Observe that the slope $m=-2h+3=3-2\sqrt{2}$. 
Since you have asked for $4$ distinct solutions, so 
$$\color{red}{0 <m < 3-2\sqrt{2}}.$$
