Critcal values of coefficient matrix with parameter where the phrase portrait changes - Boyce, p410, Question 7.6.19 The coefficient matrix contains a parameter $\alpha$. Determine the eigenvalues in terms of $\alpha$.
Find the critical value(s) of $\alpha$ where the qualitative nature of the phase portrait for the system changes.
$19. \; \mathbf{x}'(t)=\left(\begin{array}{ll}
\alpha & 10\\
-1 & -4
\end{array}\right) \mathbf{x}$
The characteristic equation for the system is given by $
r^{2}+(4-\alpha)r+10-4\alpha\ =0. $
The roots are $ r_{1,2}=-2+\dfrac{\alpha}{2}\pm\sqrt{\alpha^{2}+8\alpha-24}.
$
I use a instead of $\alpha$. When are the roots real and complex? The discriminant $= \alpha^{2}+8\alpha-24 = 0 \iff ... \iff a = -4 \pm 2\sqrt{10} $.
Because this is a parabola with a positive leading coefficient, it's concave upwards. Hence the roots are real when $a < -4 - 2\sqrt{10}$ and $ a > -4 + 2\sqrt{10} $. I color this orange.
They're complex when $-4 - 2\sqrt{10} < a <  -4 + 2\sqrt{10}  $. 
The roots change sign $\iff r = 0 \iff -2 + a/2 = \mp \sqrt{\alpha^{2}+8\alpha-24} \iff ... \iff a = 5/2$.
I want to know when the roots $<,> 0$, hence I test with some convenient numbers. WHen $a = 4$, then $r = -2 + 2 \pm \sqrt{...} $. 
Hence when $a > 5/2$, the roots have opposite signs. When $a < 5/2$, the roots have the same signs.


*

*$\color{tomato}{\alpha <-4-2\sqrt{10}}$: both roots are negative, with the equilibrium point being a stable node.

*$-4-2\sqrt{10}<\alpha <  \color{red}{2} $: the equilibrium point is a stable spiral.

*$\alpha =2$: the equilibrium point is a center.

*When $  \color{red}{2} <\alpha <-4+2\sqrt{10}$: the equilibrium point is an unstable spiral. 

*$\color{tomato}{-4+2\sqrt{10}<\alpha <2.5}$, the roots are both positive, and the equilibrium point is an unstable node. 

*$\color{tomato}{a >2.5}$ : the roots have opposite signs, with the equilibrium point being a saddle.



But where does the $ \color{red}{2} $ come from for intervals 2 to 4, when the roots are complex? I think my work so far only proves 1, 5, 6? Can someone please check my work? 

 A: If the book you're referring to is Elementary Differential Equations and Boundary Value Problems, by William E. Boyce and Richard C. DiPrima, I have checked the $10^{th}$ edition and the only reported solution is actually what you found (there is no $\alpha=2$ case anywhere).
Nonetheless, there are some mistakes here and there. Let $\Delta = \alpha^2 +8\alpha -24$. First of all, the roots are
$$
r_{1,2} = \frac{\alpha - 4 \pm \sqrt{\Delta}}{2} = \frac{\alpha}{2} - 2 \pm \frac{1}{2}\sqrt{\alpha^2 +8\alpha -24}
$$
The $1/2$ is missing in your equation. Fortunately, it doesn't change the critical values of $a_1 = -4-2\sqrt{10}$ and $a_2 = -4 +2\sqrt{10}$, which are correct, and also $a_3 = 5/2$ is correct.
However, there is a difference in the study of the positivity of the roots. Let
$$
r_1 = \frac{\alpha -4 - \sqrt{\Delta}}{2}, \quad r_2 = \frac{\alpha -4 +\sqrt{\Delta}}{2}
$$
Under the assumption that $\Delta > 0$, when you study $r_1 > 0$, you must solve $\alpha - 4 > \sqrt{\Delta}$, from which you get the following system of equations
$$
\begin{cases}
\Delta > 0 \\
\alpha -4 > 0 \\
\alpha^2 +8\alpha -24 < (\alpha -4)^2 \\
\end{cases}
$$
which has no solution. Thus, $\Delta > 0 \Longrightarrow r_1 < 0$ and checking the cases when $\Delta = 0$ (i.e., $\alpha = -4 -2\sqrt{10}$ and $\alpha = -4+2\sqrt{10}$), you can easily prove that $\Delta \geq 0 \Longrightarrow r_1 < 0$. Observe that, when $\Delta=0$, $r_1=r_2<0$, i.e., both eigenvalues are strictly negative.
Following a similar reasoning, you get that $r_2$ is positive only for $\alpha > 5/2$ ($r_2=0$ for $\alpha=5/2$), and negative for $-4+2\sqrt{10} \leq \alpha < 5/2$ and also negative for $\alpha \leq -4-2\sqrt{10}$. I leave you to fill the details, but the computations are easy.
There could have been an additional case when both eigenvalues are complex: this is the case when the real part of both roots is equal to zero (in which case, the equilibrium is a center). However, this cannot happen in your problem.
