[Math] For which value of $k$ does the matrix $A$ have one real eigenvalue of multiplicity $2$

Question

For which value of $k$ does the matrix $$A = \begin{bmatrix}-6&k\\-1&-2\end{bmatrix}$$ have one real eigenvalue of multiplicity $2$?

So I understand that if the discriminant is $0$ such that $b^2 – 4ac = 0$ then we have a root of multiplicity of 2 .

So I did $64-48+4k \implies k = -4$ and for some reason it keeps saying I'm incorrect. What am I doing wrong.

Answer 1

Note that the characteristic polynomial of $ A= \left[\begin{array}{rr} -6 & k \\ -1 & -2 \end{array}\right] $ is $$ \chi_A(t) =\det(tI-A) =\det \left[\begin{array}{rr} t + 6 & -k \\ 1 & t + 2 \end{array}\right] =t^{2} + 8 \,t + k + 12 $$ This polynomial has a real repeated root if and only if $$ 8^2-4(k+12)=0 $$ Hence $A$ has one real eigenvalue with algebraic multiplicity two if and only if $k=4$.