Prove that $\det(A^2 + A + xI) = x$

Let $x > 0$ and $A \in \mathbb R^{2 \times 2}$ satisfy $\det(A^2 + xI) = 0$. Prove that $$\det(A^2 + A + xI) = x$$

I have tried something with characteristic polynomial and eigenvalues but it did not work. Can you give me a hint to solve this problem?

Note that $$ \det(A^2 + xI) = \det(A + i\sqrt x I)\det(A - i\sqrt x I) $$ Since $A$ is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that $A$ has eigenvalues $\pm i \sqrt x$.

Now, if $\lambda$ is an eigenvalue of $A$, then $\lambda^2 + \lambda + x$ is an eigenvalue of $A^2 + A + xI$. Thus, the matrix $A^2 + A + xI$ has eigenvalues $$ (i\sqrt x)^2 + i\sqrt x + x = i\sqrt x, \\ (-i\sqrt x)^2 - i\sqrt x + x = -i\sqrt x $$ Now, $\det(A^2 + A + xI)$ is the product of these eigenvalues, which is to say $$ \det(A^2 + A + xI) = (i\sqrt{x})(-i\sqrt{x}) = x $$ as desired.

Alternatively: after finding the eigenvalues of $A$, deduce that $$ A^2 + xI = (A + i\sqrt{x} I)(A - i\sqrt{x} I) = 0 $$ (by Cayley-Hamilton), so that $$ \det(A^2 + A + xI) = \det(0 + A) = \det(A) = (i\sqrt x)(-i\sqrt x) = x $$

Let $\lambda_{1,2}$ be the eigenvalues of $A$. Then $\lambda_{1,2}^2$ are the eigenvalues of $A^2$, and hence $-\lambda_{1,2}^2$ are the eigenvalues of $-A^2$

As $\det(xI-(-A^2))=0$, it follows that $x$ is an eigenvalue of $-A^2$.

Therefore $x=-\lambda_j^2$ for $j=1$ or $j=2$. Without loss of generality $x=-\lambda_1^2$. This implies that $\lambda_1$ is purely complex, and as $A$ is a real matrix, $\lambda_2$ is the conjugate of $\lambda_1$. From $x=-\lambda_1^2$ you then get $$\lambda_{1,2} = \pm i \sqrt{x}$$

This implies that the eigenvalues of $A^2+A+xI$ are $ -x \pm i \sqrt{x} +x = \pm i \sqrt{x}$. The determinant conclusion follows.