Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$

According to the Vieta's formulas and the proposed relation, one has that
\begin{align*}
\begin{cases}
x_{1} + x_{2} + x_{3} = 3\\\\
2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = k\\\\
2x_{1}x_{2}x_{3} = 1\\\\
x^{2}_{1} + x^{2}_{2} + x^{2}_{3} = 6
\end{cases}
\end{align*}
Consequently, we have that
\begin{align*}
3^{2} = (x_{1} + x_{2} + x_{3})^{2} = x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = 6 + k \Rightarrow \boxed{k = 3}
\end{align*}
