2002 AMC 10B Problem 10

1. A quadratic that starts with $x^2$ and has solutions $a$ and $b$ can be written as $(x - a)(x - b)$. Multiply it out:
$$(x - a)(x - b) = x^2 - (a + b)x + ab.$$
2. This must be the same as $x^2 + ax + b$. Match the parts:
- The $x$ terms: $-(a + b) = a$, which means $a + b = -a$.
- The number terms: $ab = b$.
3. Start with $ab = b$. The problem says $b$ is not zero, so we can divide both sides by $b$ and get $a = 1$.
4. Put $a = 1$ into $a + b = -a$: $1 + b = -1$, so $b = -2$.
5. Check: $x^2 + x - 2 = (x - 1)(x + 2)$, whose solutions are $1$ and $-2$. Those are $a$ and $b$.

The answer is $\boxed{\textbf{(C)}\ (1,-2)}$.

For a quadratic starting with $x^2$, the solutions add to the negative of the $x$ coefficient and multiply to the constant term. These facts turn "these numbers are the solutions" into two short equations, with no quadratic formula needed. Watch for "nonzero" in a problem: it is usually the permission you need to divide.

Use the choices. From step 2, $ab = b$ with $b$ not zero forces $a = 1$. Only choice (C) has $a = 1$, and it passes the check $a + b = 1 + (-2) = -1 = -a$.

For a monic quadratic $x^2 + ax + b$, the sum of the roots is $-a$ and the product is $b$. Here the roots are $a$ and $b$ themselves, so
$$a + b = -a, \qquad ab = b.$$

Because $b \ne 0$, dividing the second equation by $b$ gives $a = 1$. Substituting into the first: $1 + b = -1$, so $b = -2$.

Check: $x^2 + x - 2 = (x+2)(x-1)$ has roots $1$ and $-2$, which are exactly $a$ and $b$.

The answer is $\boxed{\textbf{(C)}\ (1,-2)}$.