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?

If $f(x) = x^2-(3+m)x+2$, the discriminant is
$D = (3+m)^2-4\cdot 2 = (m+3)^2-8.$
This gives you a quadratic polynomial in $m$. If you have two solutions for $m$, you get in total four solutions of $f(x)$.
