<s>[INST] Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation 
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form 
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?
 [/INST] For those amongst you folks, like egreg and myself, who enjoy completing the square, we can also have at it like this:
Set
$J = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}; \tag 1$
then our equation reads
$X^2 + X = J, \tag 2$
from which we may write, completing the square, 
$(X + \dfrac{1}{2}I)^2 = X^2 + X + \dfrac{1}{4} = J + \dfrac{1}{4}I = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 3$
our task is now to find the matrices $C$ such that
$C^2 = J + \dfrac{1}{4}I = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 4$
to this end we observe that, setting
$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \tag 5$
whence
$P^2 = I, \tag 6$
we have
$\left (\dfrac{1}{2}I + P \right )^2 = \dfrac{1}{4} I^2 + 2\left ( \dfrac{1}{2}I \right ) P + P^2 = \dfrac{1}{4} I + I + P =  \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}, \tag 7$
and also
$\left (I + \dfrac{1}{2} P \right )^2 = I^2 + 2I \left (\dfrac{1}{2}P \right ) + \dfrac{1}{4} P^2 = I + \dfrac{1}{4}I + P =  \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 8$
we are thus motivated to generalize by seeking solutions of the form $aI + bP$:
$(aI + bP)^2 = a^2 I^2 + 2abIP + b^2 P^2 = a^2 I + b^2 I + 2ab P$
$= (a^2 + b^2)I + 2ab P =  \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix} = \dfrac{5}{4}I + P; \tag 9$
thus,
$a^2 + b^2 = \dfrac{5}{4}; \; 2ab = 1; \tag{10}$
it follows from (10) that the ordered pairs $(a, b)$ lie on both the circle of radius $5/4$ centered at the origin in the $(a, b)$ plane, and also on the hypberbola $ab = 1/2$; since the points $\pm (1/\sqrt 2, 1/\sqrt 2)$ lie in the interior of the circle, the branches of this hyperbola cross the circle at exactly four points, and we have seen in (8)-(9) that two of them are $(1, 1/2)$ and $(1/2, 1)$; it is then evident from the symmetry $(a, b) \leftrightarrow (-a, -b)$ of the solutions to (10) that there are a total of precisely four pairs $(a, b)$ which solve (10) and hence (9), viz.
$(a, b) = \pm \left (1, \dfrac{1}{2} \right ), \; \pm \left ( \dfrac{1}{2}, 1 \right ); \tag{11}$
since
$X + \dfrac{1}{2}I = aI + bP = \pm \begin{bmatrix} 1 & \dfrac{1}{2} \\ \dfrac{1}{2} & 1 \end{bmatrix}, \pm \begin{bmatrix} \dfrac{1}{2} & 1 \\ 1 & \dfrac{1}{2} \end{bmatrix}, \tag{12}$
we have
$X = \begin{bmatrix}  \dfrac{1}{2} & \dfrac{1}{2} \\ \dfrac{1}{2} &  \dfrac{1}{2} \end{bmatrix}, \; \begin{bmatrix} -1 & -1 \\ -1 & -1  \end{bmatrix}, \; \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} -\dfrac{3}{2} & -\dfrac{1}{2} \\ -\dfrac{1}{2} & -\dfrac{3}{2} \end{bmatrix}, \tag{13}$
as the solutions to (2).
One question does, however, remain:  how do we know every $C$ as in (4) may be cast in the form $aI + bP$?  Well, as pointed out by Lord Shark the Unknown in his answer, $XJ = JX$ implies
$X = aI + bJ; \tag{14}$
but
$J = I + P, \tag{15}$
so in fact
$a I + b J = aI + b(I + P) = (a + b)I + bP, \tag{16}$
which shows that $X$ may be represented in the form $aI + bP$ as well; thus so may
$C = X + \dfrac{1}{2}I = (a + \dfrac{1}{2})I + bJ = (a + b + \dfrac{1}{2})I + bP. \tag{17}$
</s>