Finding $k$ such that a given matrix has a real eigenvalue of algebraic multiplicity $2$ I am trying to find, for which values $k$, the matrix below has a real eigenvalue with algebraic multiplicity $2$:

\begin{pmatrix}
-3 &amp; k\\ 
2 &amp; -6
\end{pmatrix}

My work thus far:

$$(-3-λ)(-6-λ)-2K$$

$$λ^2 +9λ+18-2k$$

$$\frac{-9±⌈9-8k⌉}{2}$$

I'm not sure where to go on from here to solve this.
For us to have a double root (i.e. an algebraic multiplicity of two), the discriminant of the characteristic equation must be zero. This holds for quadratics in general: if $f(x) = ax^2 + bx + c$ has a root of multiplicity two - a double root - then $b^2 - 4ac = 0$.



The characteristic equation in this case is

$$\lambda^2 + 9\lambda + (18-2k) = 0$$

and the discriminant being the rooted expression in the quadratic formula, which, here, is

$$9^2 - 4(1)(18-2k) = 81 - (72-8k) = 9 + 8k$$

So, for the discriminant to be zero, we require

$$9+8k = 0$$

Only one $k$ satisfies this equation, and I imagine you can find it pretty easily.
For us to have a double root (i.e. an algebraic multiplicity of two), the discriminant of the characteristic equation must be zero. This holds for quadratics in general: if $f(x) = ax^2 + bx + c$ has a root of multiplicity two - a double root - then $b^2 - 4ac = 0$.



The characteristic equation in this case is

$$\lambda^2 + 9\lambda + (18-2k) = 0$$

and the discriminant being the rooted expression in the quadratic formula, which, here, is

$$9^2 - 4(1)(18-2k) = 81 - (72-8k) = 9 + 8k$$

So, for the discriminant to be zero, we require

$$9+8k = 0$$

Only one $k$ satisfies this equation, and I imagine you can find it pretty easily.