<s>[INST] 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
$$x_1^2+x_2^2+x_3^2 = 6$$
 [/INST] Rearrange $2x^3 - 6x^2 + kx -1=0$ as $x(2x^2+k)=6x^2+1$, and square both sides to get the equation in $x^2$
$$x^6 +(k-9)x^4+(\frac{k^2}4-3)x^2-\frac14=0$$
Then
$$x_1^2+x_2^2+x_3^2= 6= -(k-9)$$
which yields $k=3$.
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