11.45 For what values of $m$ does the equation $2 \sqrt{1-m(x+2)}=x+4$ have one root?
11.45 If \( m=0 \), then the equation will take the form \( 2=x+4 \) and, therefore, has a single root \( x=-2 \). Let \( m \neq 0 \). We set

\[
\sqrt{1-m(x+2)}=y \geq 0, \quad \text { from which after }
\]

transformations we find \( x=\frac{1-y^{2}}{m}-2 \). Then the given equation can be written as:

\[
2 y=\frac{1-y^{2}}{m}-2+4, \text { or } y^{2}+2 m y-1=2 m
\]

Now let's construct the graphs of the functions \( z=y^{2}+2 m y-1 \) and \( z=2 m \). Since the coefficient of \( y \) in the parabola equation contains the parameter \( m \), we need to consider two cases: \( m>0 \) and \( m<0 \).

If \( m>0 \), then the line \( z=2 m \) intersects the specified part of the parabola at a single point under the condition \( 2 m>-1 \), i.e., when \( m>-\frac{1}{2} \); hence, in this case \( m>0 \).

If \( m<0 \), then the equation has a single root under the condition that the line \( z=2 m \) is tangent to the parabola \( z=y^{2}+2 m y-1 \) at its vertex. Writing the equation of the parabola as \( z=(y+m)^{2}-1-m^{2} \), we establish that the ordinate of the vertex is \( -1-m^{2} \); thus, the tangency condition becomes \( -1-m^{2}=2 m \), from which \( m=-1 \). Therefore, combining all the obtained information, we conclude that the equation has a single root when \( m=-1 \) and when \( -\frac{1}{2}<m<\infty \).