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.

The parabola $x^2 + 4x + 3 = (x+3)(x+1)$ is upward facing and has roots at $x = -3, x = -1$, so it is negative in $(-3, -1)$. Break it into cases. One of these cases has two solutions, the other has exactly one (i.e. discriminant = 0).
Case 1: $-3 \leq x \leq -1$. The equation is $-x^2 - 4x - 3 - mx + 2m = 0$ so $x^2 + (4+m)x + (3-2m) = 0$. Discriminant is $\sqrt{(m+4)^2 - 4(3-2m)}$. 
Case 2: The other case. The equation is $x^2 + (4-m) x + (3+2m) = 0$, with discriminant $\sqrt{(4-m)^2 - 4(3+2m)}$. 
Suppose Case 1 has one solution and Case 2 has two. Then $m^2 + 8m + 16 - 12 + 8m = 0 \to m^2 + 16m + 4 = 0$ so $m = -8 \pm 2 \sqrt{15}$. The root of the quadratic is $\frac{-(m+4)}{2}$, and so $m = -8 + 2 \sqrt{15}$ works for the domain. This also gives valid roots in the other quadratic, so this is our solution.
Remark. If we instead suppose that Case 2 has the one solution we end up finding two values of $m$ which give invalid solutions for the Case 1 quadratic.
