Suppose that $a$ and $b$ are nonzero real numbers, and that the [equation](https://artofproblemsolving.com/wiki/index.php/Equation) $x^2 + ax + b = 0$ has solutions $a$ and $b$. Then the pair $(a,b)$ is
$\mathrm{(A)}\ (-2,1) \qquad\mathrm{(B)}\ (-1,2) \qquad\mathrm{(C)}\ (1,-2) \qquad\mathrm{(D)}\ (2,-1) \qquad\mathrm{(E)}\ (4,4)$
Since $(x-a)(x-b) = x^2 - (a+b)x + ab = x^2 + ax + b = 0$, it follows by comparing [coefficients](https://artofproblemsolving.com/wiki/index.php/Coefficient) that $-a - b = a$ and that $ab = b$. Since $b$ is nonzero, $a = 1$, and $-1 - b = 1 \Longrightarrow b = -2$. Thus $(a,b) = \boxed{\mathrm{(C)}\ (1,-2)}$.