Real $2\times 2$ matrix $X$ such that $X^2 + 2X= -5I$ Find a real $2\times 2$ matrix $X = \left(\begin{matrix} a&amp; b\\ c &amp; d\end{matrix}\right)$ 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.
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
$$X=\begin{pmatrix}-1&amp;4\\-1&amp;-1\end{pmatrix}$$
Expanded you get that

\begin{matrix}
        5 + a (2 + a) + b c &amp; b (2 + a + d) \\
        c (2 + a + d) &amp; 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 -&gt; -1 - 2i, b -&gt; 0, c -&gt; 0, d -&gt; -1 - 2i}, 
{a -&gt; -1 - 2i, b -&gt; 0, c -&gt; 0, d -&gt; -1 + 2i},
{a -&gt; -1 + 2i, b -&gt; 0, c -&gt; 0, d -&gt; -1 - 2i}, 
{a -&gt; -1 + 2i, b -&gt; 0, c -&gt; 0, d -&gt; -1 + 2i}