range of $m$ such that the equation $|x^2-3x+2|=mx$ has 4 real answers. Find range of $m$ such that the equation $|x^2-3x+2|=mx$ has 4 distinct real solutions $\alpha,\beta,\gamma,\delta$
To show how I got the wrong answers.
From $|x^2-3x+2|=mx$
I got the two case $x^2-3x+2=mx$ when $x>2 $ or $ x<1$
and $x^2-3x+2=-mx$ when $1<x<2$
also $m\neq0$ (because if $m=0$ , This will given only 2 answers not 4)
try to find the first two answers $x^2-3x+2=mx$ when $x>2 $ or $ x<1$
$x^2-(3+m)x+2=0$  when $x>2 $ or $ x<1$
Use quadratic formula will given $x= \frac{3+m\pm \sqrt{(3+m)^2-4\times2}}{2}$
$x$ will be real number and have two answers if $\sqrt{(3+m)^2-4\times2} > 0$
$m^2+6m+11>0$, got that $-3-2\sqrt{2}<m<-3+2\sqrt{2}$
and on the another case where $x^2-3x+2=-mx$ when $1<x<2$
$x^2-(3-m)x+2=0$ when $1<x<2$
Use quadratic formula will given $x= \frac{3-m\pm \sqrt{(3-m)^2-4\times2}}{2}$
$x$ will be real number and have the others two answers if $\sqrt{(3-m)^2-4\times2} > 0$
$\sqrt{(3-m)^2-4\times2} > 0$, got that $3-2\sqrt{2}<m<3+2\sqrt{2}$
So, I believe that the answers should be $(-3-2\sqrt{2}<m<-3+2\sqrt{2}) \cup (3-2\sqrt{2}<m<3+2\sqrt{2})$
However, the book's right answer is $0 < m < 3-2\sqrt{2}$
Please show me the method to obtain the right answers.

There is some positive value $m$ such that $y=mx$ is tangent to $y=-(x^2-3x+2)$. 

This value must make $0$ the discriminant of the equation
$$x^2-3x+2=-mx$$
That is, $$m^2-6m+1=0$$
The least root of this equation is $$3-2\sqrt2$$
So $0<m< 3-2\sqrt 2$.
