Help with this problem: quadratic forms and semi-definiteness.

Question

Given $$Q(x_1,x_2) = (4+k)x_1^2 + 2kx_1x_2 + 2x_2^2$$ I'm asked to find the values of $k$ such that $Q$ be a positive semi-definite form. Now, to do so I've decided to find its characteristic polynomial $$P(\lambda) = \lambda^2 - k^2+ 6\lambda - k\lambda + 8 + 2k$$ in order to build this system of equations: $$\begin{cases} P(\lambda) = 0 \\ \lambda \geq 0 \end{cases}$$ that later leads me to a dead end or nonsensical conclusions as $$k \geq \frac{1}{2}(2 - \lambda - \sqrt{36 - 28 \lambda + 5 \lambda^2}) - \lambda $$

My biggest concern, apart from the fact that my result seems so stupid and not in accordance to what I'm given as the right answer, is that the $\lambda \geq 0$ in my system of equations seems abusive, since I have the impression it imposes that there is one and only one eigenvalue. I know that my system of equations stinks, but I don't know how to put the problem.

How should I procede to solve this exercice and, more generally, what method should I use to face this kind of problems?

Thank you in advance.

EDIT:

I found the answer to this exercice by solving the following equation, involving the coefficient matrix of the quadratic form: $$ \begin{vmatrix}4+k&k\\k&2\end{vmatrix} =-k^2+2k+8 = 0$$ leading to $$k =\begin{cases}\ \ \ 4\\-2\end{cases}$$ which is the answer given as correct. Nevertheless, is this the or a generally correct method?

Answer 1

See Sylvester's Law of Inertia

$$ \left( \begin{array}{rr} 1 & - \frac{k}{2} \\ 0 & 1 \end{array} \right) \left( \begin{array}{cc} 4+k & k \\ k & 2 \end{array} \right) \left( \begin{array}{rr} 1 & 0 \\ - \frac{k}{2} & 1 \end{array} \right) = \left( \begin{array}{cc} 4 + k - \frac{k^2}{2} & 0 \\ 0 & 2 \end{array} \right) $$

Answer 2

For $Q$ to be semidefinite, one of the eigenvalues of its matrix must be zero, i.e., it must be singular, which is why finding where its determinant vanishes as in your answer gives you possible values of $k$. You still have to verify that $Q$ is indeed positive-definite, and not negative-definite, for these values of $k$.

Relating this to your approach, for $Q$ to be positive-definite, one of the roots of the characteristic polynomial must be $0$. This gives you two equations to solve for $k$, specifically, $$\frac12(k+6\pm\sqrt{5k^2+4k+4})=0.$$ One of these has no real solutions and the solutions to the other match the ones from the determinant.