One root of $x^2-(k+3)x+8=0$ is twice the other. The positive value of $k$ is:

Take the roots as $r$ and $2r$. Their product $2r^2=8$ gives $r$ = ±2, and their sum $3r=k+3$. With $r=2$ we get $k+3=6$, so $k=3$; with $r=-2$ we get $k=-9$. The positive value is 3. (6 is the sum of the roots, $k+3$, not $k$ itself.) At $k=3$ the equation is $x^2-6x+8=0$ with roots 2 and 4.