(1 point) For which value of $k$ does the matrix
\[
A=\left[\begin{array}{cc}
-6 & k \\
6 & 8
\end{array}\right]
\]
have one real eigenvalue of multiplicity 2 ?
\[
k=
\]

Step 1 ：The eigenvalues of a matrix are the roots of the characteristic polynomial, which is given by det(A - λI) = 0, where A is the matrix, λ are the eigenvalues, I is the identity matrix, and det denotes the determinant. In this case, we have: \[A - λI = \left[\begin{array}{cc}-6 - λ & k \\6 & 8 - λ\end{array}\right]\]

Step 2 ：The characteristic polynomial is (-6 - λ)(8 - λ) - 6k = λ^2 + 2λ + 48 - k = 0.

Step 3 ：For the matrix to have one real eigenvalue of multiplicity 2, the discriminant of this quadratic equation must be zero. The discriminant is given by b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation. In this case, a = 1, b = 2, and c = 48 - k.

Step 4 ：We need to solve the equation 2^2 - 4(1)(48 - k) = 0 for k.

Step 5 ：The value of $k$ for which the matrix has one real eigenvalue of multiplicity 2 is \(\boxed{47}\).