11. $x^{3}+k x-128=0$ has a root of multiplicity 2 . Find $k$.
(ans. -48
$$
\begin{array}{l}
x^{3}+k x-128=(x-r)^{2}(x-s)=x^{3}+(s-2 r) x^{2}+\left(r^{2}+2 r s\right) x-r^{2} s= \\
0 \Rightarrow r^{s}=128, s+2 r=0 \Rightarrow s=-2 r,-2 r^{3}=128 \Rightarrow r=-4, s=8 .
\end{array}
$$
$$
\text { Thus, } k=r^{2}+2 r s=16-64=-48 \text {.) }
$$