2013 AMC 10B Problem 19

An arithmetic sequence $c, b, a$ means $b$ is the average: $b = \frac{a + c}{2}$. Exactly one root means the discriminant vanishes: $b^2 = 4ac$. Substituting,
$$\frac{(a + c)^2}{4} = 4ac \quad\Longrightarrow\quad a^2 - 14ac + c^2 = 0.$$
Here $a > 0$ (otherwise everything is $0$ and there is no quadratic), so divide by $a^2$ and set $t = \frac{c}{a}$:
$$t^2 - 14t + 1 = 0 \quad\Longrightarrow\quad t = 7 \pm 4\sqrt3.$$
Since $c \le a$, we need $t \le 1$, so $t = 7 - 4\sqrt3$.

The double root of $ax^2 + bx + c$ is $x = -\frac{b}{2a}$. With $b = \frac{a + c}{2}$,
$$x = -\frac{a + c}{4a} = -\frac{1 + t}{4} = -\frac{8 - 4\sqrt3}{4} = -2 + \sqrt3.$$

The answer is $\boxed{\textbf{(D)}\ -2+\sqrt{3}}$.

Both conditions are homogeneous in $a, b, c$, so only the ratios matter; dividing by $a^2$ reduces three unknowns to one. The ordering $a \ge c$ is what selects the correct sign of the square root, and the root of a perfect-square quadratic is read off as $-\frac{b}{2a}$ without solving anything further.

Check the choices: the root of a quadratic with all nonnegative coefficients must be negative and, since $a \ge b \ge c$, cannot be too far from $-\frac12$ in size; $-2 + \sqrt3 \approx -0.27$ fits while $-7 + 4\sqrt3 \approx -0.07$ and the choices below $-1$ do not. Confirm with $a = 1$, $c = 7 - 4\sqrt3$, $b = 4 - 2\sqrt3$: then $b^2 = 28 - 16\sqrt3 = 4c$.

• Finding the ratio a/c = 7 + 4sqrt3 and reporting it (or its negative) instead of the root -b/(2a).
• Choosing $t = 7 + 4\sqrt3$ (which violates $c \le a$) and getting the root $-2 - \sqrt3$, choice (B).