If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem : 
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$. 
My Approach: 
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0  $
Discriminant of above qudratic is 
$D = (4-m)^2 -4(3+2m) \geq 0$
$D = 16+m^2-8m-12-8m$
Solving for $m$ we get the values $-8 \pm 2\sqrt{15}$
Case II : 
Similarly solving for the given equation taking negative sign of modulus we get the solution 
for $m =$$8 \pm 2\sqrt{15}$ 
Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.

$$m(x-2)=|(x+3)(x+1)|\ge0$$
If $m=0,$ there are two real solutions
Else $m(x-2)=|(x+3)(x+1)|=0$ has no solution
So, $$m(x-2)=|(x+3)(x+1)|>0$$
Now $|(x+3)(x+1)|=-(x+3)(x+1)$ if $-3\le x\le-1$
$=+(x+3)(x+1)$ otherwise
If $m>0,x-2>0\iff x>2\implies m(x-2)=x^2+4x+3$ which has exactly two solutions
If $m<0,x-2<0\iff x<2$
If $-1<x<2$ or if $x<-3;$   $m(x-2)=(x+3)(x+1)\  \  \  \ (1)$
If $-3\le x\le-1,m(x-2)=-(x+3)(x+1)\  \  \  \ (2)$
We need the discriminant of $(1)$ or $(2)$ to be zero honoring the range.
