Matrix $A$ has two distinct real eigenvalues iff $k$ is greater than what? 
The matrix $A = \begin{bmatrix}3&k\\8&8\end{bmatrix}$ has two distinct real eigenvalues iff $k > ?$

So I found the determinant by doing:
$(3 - \lambda)(8 - \lambda) - 8k = \lambda^2 - 11\lambda + 24 - 8k \implies \lambda = 8, \lambda = 3$ The thing is, I'm not really sure what they are asking me because I have found what the eigenvalues are: $\lambda_1 = 8, \lambda_2 = 3$. 
I'm assuming I need to solve for $k$ somehow but it doesn't seem very straightforward to me, what am I missing here?
 A: Note that the characteristic equation of this matrix is given by
$$
\lambda^2 - 11\lambda + (24 - 8k) = 0
$$
The question is to find which values of $k$ produce two distinct real roots $\lambda$ to this equation.  It is helpful to consider the discriminant of this quadratic equation, namely
$$
b^2 - 4ac = 121 - 4(24 - 8k)
$$
recall that a quadratic equation will have distinct real roots iff its discriminant is positive.
A: The determinant is $$(3 - \lambda)(8 - \lambda) - 8k = \lambda^2 - 11\lambda + 24 - 8k.$$
In order to find the eigenvalues, you have to solve the equation $$\lambda^2 - 11\lambda + 24 - 8k = 0.$$
The solutions are:
$$\lambda = \frac{11 \pm \sqrt{11^2 - 4(24-8k)}}{2} = \frac{11 \pm \sqrt{25+32k}}{2}$$
which of course depend on $k$.
The two solutions are different and real if the part under square root is positive. That is:
$$25 +32 k > 0 \Rightarrow k > -\frac{25}{32}.$$
A: $\lambda^2 - 11\lambda + 24 - 8k = 0\\
\lambda = \dfrac {11 \pm \sqrt{25 +32k}}{2} $
$A$ has 2 distinct real egeinvalues if
$25 +32k>0$
A: You need to solve the equation with $\lambda$ the variable and $k$ a parameter. 
The solutions are 
$$ -\frac{11}{2} \pm\sqrt{\frac{11^2}{4} - (24 - 8k)  } $$
These are the eigenvalues of the matrix in dependence of $k$. 
These two are distinct and real  if and only if 
$$\frac{11^2}{4} - (24 - 8k)> 0.$$ 
Solve this to get your  $k$.
