Any real quartic has an even number of real roots with multiplicity, so there exists real $r$ such that $x^{4}+k x^{3}+x^{2}+4 k x+16$ either takes the form $(x+r)^{4}$ (clearly impossible) or $(x+r)^{2}\left(x^{2}+a x+b\right)$ for some real $a, b$ with $a^{2}<4 b$. Clearly $r \neq 0$, so $b=\frac{16}{r^{2}}$ and $4 k=4(k)$ yields $\frac{32}{r}+a r^{2}=4(2 r+a) \Longrightarrow a\left(r^{2}-4\right)=8 \frac{r^{2}-4}{r}$. Yet $a \neq \frac{8}{r}$ (or else $a^{2}=4 b$ ), so $r^{2}=4$, and $1=r^{2}+2 r a+\frac{16}{r^{2}} \Longrightarrow a=\frac{-7}{2 r}$. Thus $k=2 r-\frac{7}{2 r}= \pm \frac{9}{4}$ (since $r= \pm 2$ ).