2008 AMC 10B Problem 9

We do not need the solutions themselves, only their average. The sum of the solutions of a quadratic can be read straight from its coefficients.

1. A quadratic can only have two solutions if $a$ is not $0$, so divide the whole equation by $a$:
$$x^2 - 2x + \frac{b}{a} = 0.$$
2. For $x^2 + px + q = 0$, the two solutions add to $-p$. (If the solutions are $r$ and $s$, the equation is $(x - r)(x - s) = x^2 - (r + s)x + rs$, so the middle coefficient is minus the sum.)
3. Here the middle coefficient is $-2$, so the solutions add to $2$.
4. Their average is $2 \div 2 = 1$. The value of $b$ only decides whether the solutions are real; it does not move their average.

The answer is $\boxed{\textbf{(A)}\ 1}$.

The two solutions of a quadratic sit evenly on either side of the middle of its graph (the vertex), so their average is the vertex's $x$-value. That depends only on the first two coefficients. Any time a question asks for the sum or average of the solutions, read it from the coefficients instead of solving.

Try numbers. With $a = 1$ and $b = 0$ the equation is $x^2 - 2x = 0$, or $x(x - 2) = 0$, with solutions $0$ and $2$. Their average is $1$. For these values the choices give (A) $1$, (B) $2$, (C) $0$, (D) $0$, (E) not a real number, so the answer is (A).