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 value of $m$ you are looking for is the slope of the tangent line to the graph $y = -x^2 - 4x _ 3, -3 < x < 1.$ it has a tangent at the $x$-coordinate $2 - \sqrt {15}$ with a slope of $-2a - 4 = -8 + 2\sqrt{15}.$ any line with a bigger slope cuts at two points and any with a positive and smaller slope will cut at four points and if the slope is zero, then it cuts at two pints $x = -1, x = -3.$
