Real $2\times 2$ matrix $X$ such that $X^2 + 2X= -5I$ Find a real $2\times 2$ matrix $X = \left(\begin{matrix} a& b\\ c & d\end{matrix}\right)$ such that $X^2 + 2X = -5I.$ 
With this question I'm kinda lost with the $2X$ part but a full explanation or a little bit of a lead would be greatly appreciated. 
 A: For a $2\times 2$ matrix $X$ its characteristic polynomial has the form
$$\chi_X(\lambda)=\lambda^2-\operatorname{tr}(X)\lambda+\det X$$
and by the Cayely-Hamilton theorem we have so
$$X^2-\operatorname{tr}(X)X+(\det X)I=0$$
so to answer the question it suffices to construct $X$ such that $\operatorname{tr}(X)=a+d=-2$ and $\det X=ad-bc=5$. Take for example
$$X=\begin{pmatrix}-1&4\\-1&-1\end{pmatrix}$$
A: Solve the quadratic
$$
z^2+2z+5=0\Longrightarrow z=-1\pm 2i.
$$
Now use the representation of complex  numbers by real $2\times 2$ matrices:
$$
 z=x+iy\longmapsto\begin{pmatrix}x&y\\-y&x\end{pmatrix}.
$$
This gives two solutions of our matrix equation:
$$
 X_1=\begin{pmatrix}-1&2\\-2&-1\end{pmatrix}\qquad\text{or}\qquad
 X_2=\begin{pmatrix}-1&-2\\2&-1\end{pmatrix}.
$$
The general real solution has the form
$X=U^{-1}X_1U$ with an arbitrary real invertible matrix $U$. In particular,
$$
 X_2=U^{-1}X_1U,\qquad\text{where}\qquad U=U^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
A: A number of answers to this question have been given but I just want to indicate the simple solution and its connection with the general theory. The question is for a matrix with minimal polynomial $x^2+2x+5$. This is irreducible over the ground field ($\mathbb{R}$). Let $V$ be a two dimensional space and $T$ a linear transform with the given polynomial as minimal polynomial. Let $v \in V$ be any vector Then we have also $Tv$ and finally
$T^2v=-5-2Tv$ writing vector coordinates with respect to the basis $\{v,Tv\}$ we see that the matrix for $T$, acts as follows,
$$\begin{pmatrix}
a&b\\c&d\end{pmatrix}\begin{pmatrix}1\\0
\end{pmatrix}= \begin{pmatrix}0\\1
\end{pmatrix}$$
and 
$$\begin{pmatrix}
a&b\\c&d\end{pmatrix}\begin{pmatrix}0\\1
\end{pmatrix}= \begin{pmatrix}-5\\-2
\end{pmatrix}$$
so we see immediately that 
$$X=
\begin{pmatrix}
0&-5\\1&-2\end{pmatrix}$$
This is called the companion matrix and we see that any solution will be conjugate to this solution. The advantage here is that we can simply read the solution off of the polynomial.
In general for an irreducible minimal polynomial of the form
$x^n+a_{n-1}x^{n-1}+\cdots +a_0=0$ we have companion matrix
$$\begin{pmatrix}
0&0&\cdots&0&-a_0\\
1&0&\cdots&0 &-a_1\\
0&1&\cdots &0&-a_1\\
\vdots&\vdots&\ddots &\vdots&\vdots\\
0&0&\cdots &1&-a_{n-1}\\
\end{pmatrix}$$
which has minimal polynomial $x^n+a_{n-1}x^{n-1}+\cdots +a_0=0$. 
A: Yes, real solutions exist for all $ bc \leq - 4$.
I wrote your matrix equation as a non-linear system of equations:
\begin{cases} a^2 + bc + 2a + 5 = 0 \\ ab + bd + 2b = 0 \\ ac + cd + 2c = 0 \\ bc + d^2 + 2d +5 = 0 \end{cases}
Which can be solved by your favorite method (I used substitution) to give:
$a = \mp \sqrt{-bc-4}-1 | b,c \in \mathbb{R}$
$d = \pm \sqrt{-bc-4}-1 | b,c \in \mathbb{R}$
A: Expanded you get that
\begin{matrix}
        5 + a (2 + a) + b c & b (2 + a + d) \\
        c (2 + a + d) & 5 + b c + d (2 + d)  \\
        \end{matrix} 
equals 0.
Solving using Mathematica 9
X = {{a, b}, {c, d}};
Id = IdentityMatrix[2];
Solve[ X.X + 2 X == -5 Id, {a, b, c, d}]

Atleast these are possible answers:
{a -> -1 - 2i, b -> 0, c -> 0, d -> -1 - 2i}, 
{a -> -1 - 2i, b -> 0, c -> 0, d -> -1 + 2i},
{a -> -1 + 2i, b -> 0, c -> 0, d -> -1 - 2i}, 
{a -> -1 + 2i, b -> 0, c -> 0, d -> -1 + 2i}

