<s>[INST] 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 =$$8 \pm 2\sqrt{15}$ 
Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.
 [/INST] If you recognize that $x^2 + 4x + 3 = (x+2)^2 - 1,$
then it can quickly be seen that the graph of that function is an
upward-opening parabola with its minimum at $x=-2$, where the value of the
function is $-1$.
Clearly for values of $x$ far from $-2$, the function is positive,
those parts of the graph of the function are preserved when
we take the absolute value, $|x^2 + 4x + 3|,$
but the part near $x=-2$ (bounded by the points where the parabola crosses
the $x$ axis to the left and right of $x=-2$) is flipped over the $x$ axis;
that is, the section of that parabola below the $x$ axis is replaced
by the section of the parabola $1-(x+2)^2$ above the $x$ axis.
So the graph of $|x^2 + 4x + 3|$ starts at large positive $y$ and large
negative $x$, goes down to the $x$ axis, then goes up over a "hill" 
and down again until it arrives at the $x$ axis again, and finally goes up
indefinitely.
The solutions of $|x^2+4x+3|-mx+2m=0$ are the points where
$$|x^2 + 4x + 3| = m(x - 2),$$ so we can look for a line that intercepts the
$x$ axis at $x=2$ with slope $m$ and that intersects the graph of
$|x^2 + 4x + 3|$ in exactly three points.
It is clear that a positive $m$ will not work, and for zero or negative $m$
it is easy to get either two intersection points or four;
but the only way to get three intersection points is for the line to be just
tangent to the "hill" in the middle of the graph of $|x^2 + 4x + 3|$.
(A slightly shallower slope gives four intersections; steeper gives two.)
So you need there to be a unique solution to $m(x-2) = 1-(x+2)^2$.
Rather than do much more algebra on that form of the equation,
we can observe that $1-(x+2)^2$ describes the part of the graph of
$|x^2 + 4x + 3|$ where the $|\cdot|$ operation changed the sign of the formula,
so we really want there to be one (double root) solution of
$$-(x^2+4x+3)-mx+2m = -x^2 -(4+m)x - (3-2m)=0.$$
Using the $b^2-4ac$ formula for the discriminant,
the discriminant is $(m+4)^2 - 4(3-2m)$
or simplified, $m^2 + 16m + 4$.
This has two possible solutions:
the larger (steeper) value of $m$ corresponds to a line that
is tangent to the parabola $-(x^2+4x+3)$ somewhere to the right of $x=2$
and far below the $x$-axis, in other words, nowhere near the graph of
$|x^2 + 4x + 3|$;
the smaller (shallower) slope $m$
is the solution you want.
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