2002 AMC 10B Problem 10

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

Vieta's formulas translate "these numbers are the roots" into two simple equations in the coefficients, with no need to solve the quadratic. The nonzero hypothesis is the key that unlocks $ab = b \Rightarrow a = 1$; without it, $b = 0$ would also be allowed.

Use the choices: $ab = b$ forces $a = 1$, and only (C) has first coordinate $1$. Confirm with $a + b = 1 - 2 = -1 = -a$.

• Plugging x = a and x = b into the equation and grinding through a quartic instead of reading off sum and product with Vieta.
• Writing the sum of roots as $+a$ instead of $-a$, which produces $b = -2a$ with the wrong sign and points to (B) or (D).