Given a Lie Algebra $L$ on a field $F$, we define the radical of $L$ $\operatorname{Rad}(L)$ as the largest solvable ideal of $L$.
We define the adjoint representation $\operatorname{ad}_L:L\to\mathfrak{gl}(L)$, from which we get the killing form: a bilinear symmetric form $k:L\times L\to F$ defined as $k(x,y):=\operatorname{Tr}(\operatorname{ad}_L(x) \operatorname{ad}_L(y))$.
We next define the radical of $k$ as $\operatorname{Rad}(k):=\{x\in L\;:\;k(x,L)=0\}$.
How can I prove that $\operatorname{Rad}(k)=\operatorname{Rad}(L)$?
This is not true in general. Here is a counterexample. Let $L$ be the $2$-dimensional nonabelian Lie algebra with basis $(x,y)$ and Lie brackets $[x,y]=y$. Of course, ${\rm Rad}(L)=L$, because $L$ is solvable. However, the matrix of the Killing form $\kappa$ is given by $$ \kappa = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, $$ so that ${\rm Rad}(\kappa)\neq L$; in fact, $\kappa(x,L)\neq 0$, because ${\rm tr} \, ad(x)^2=1$.
However, for a semisimple Lie algebra $L$ of characteristic zero we have, by definition, ${\rm Rad}(L)=0$. Since the Killing form of a semisimple Lie algebra is non-degenerated, we also have ${\rm Rad}(\kappa)=0$. So in this case we trivially have ${\rm Rad}(L)={\rm Rad}(\kappa)$.