Given the system $\ddot{x} +B \dot{x}+ C x =0$ where $x \in R^{2}$ and $B$ is positive definite symmetric matrix and I know that the eigenvalues of $C$ are either $\lambda_1<0<\lambda_2$ or $\lambda=\pm \omega j$.
My question is: Can I conclude from these givens that the system is unstable and can the type the equilibrium point be determined?
Let us write $B = \begin{bmatrix} b_1 & b_2 \\ b_2 & b_3\end{bmatrix}$ and $C = \begin{bmatrix} c_1 & c_2 \\ c_3 & c_4\end{bmatrix}$. Since $B$ is postive definite, we know that $b_1>0$, $b_3>0$, and $\det B = b_1b_3 - b_2^2 >0$. For $C$ we know that either
Let us define $\eta := [x_1, x_2, \dot{x}_1, \dot{x}_2]^\top$. Then your system is given by $\dot{\eta} = F\eta$, where $$F := \begin{bmatrix}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -c_1 & -c_2 & -b_1 & -b_2 \\ -c_3& -c_4& -b_2& -b_3\end{bmatrix}.$$
The characteristic polynomial is given by $$\det(sI-F) = s^4+\left(b_{1}+b_{3}\right)\,s^3+\left(-{b_{2}}^2+c_{1}+c_{4}+b_{1}\,b_{3}\right)\,s^2+\left(b_{3}\,c_{1}-b_{2}\,c_{2}+b_{1}\,c_{4}-b_{2}\,c_{3}\right)\,s+c_{1}\,c_{4}-c_{2}\,c_{3}.$$
So, in the $\lambda_1 \lambda_2<0$ case we have that the free term of the polynomial is negative, and the system is unstable.
For the second part, the answer is negative. Try (a random search result) $$B = \begin{bmatrix} 3.3 & -1 \\ -1 & 8.4 \end{bmatrix}, \quad C = \begin{bmatrix} 4.8 & 3.3 \\ -9.68 & -4.8 \end{bmatrix}.$$ The matrix $B$ is positive-definite, the matrix $C$ has eigenvalues on the imaginary axes, and $F$ is Hurwitz.