If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then the value of m is equal to to ?
I drew the graph of $|x^2+4x+3|$.I found that that for the given condition $mx-2m$ must be a tangent to $-(x^2+4x+3)$.So using the concept of putting the discriminant to $0$, I'm getting $m=-8+2\sqrt{15}$ and $m=-8-2\sqrt{15}$. But the solution $m=-8-2\sqrt{15}$ is not given in my book.I can't understand why.Only the first solution $m=-8+2\sqrt{15}$ is given.
Where am I going wrong?
Denote $p(x) := x^2 + 4 x + 3$. Then, the condition that $m x - 2 m$ be tangent to the graph of $-p(x)$ arises from the fact that the latter function coincides with $|p(x)|$ between the roots of $p$, namely, on the interval $[-3, -1]$.
Now, when $m = -8 \pm 2\sqrt{15}$, computing gives that the graph of $-p(x)$ is tangent to $m x - 2 m$ at $2 \mp \sqrt{15}$. But of these $x$-values, only $-2 + \sqrt{15}$ is in the interval, so the other solution (and hence the corresponding $m$-value, $-8 - 2 \sqrt{15}$), does not correspond to a point of tangency of the line to the graph of $|p(x)|$.