2013 AMC 10B Problem 19

1. The numbers $c, b, a$ form an arithmetic sequence, so $b$ is exactly halfway between $a$ and $c$:
$$b = \frac{a + c}{2}.$$
2. A quadratic has exactly one root when the part under the square root in the quadratic formula is zero: $b^2 - 4ac = 0$. Then the formula gives the root as $x = -\frac{b}{2a}$.
3. The number $a$ cannot be $0$: since $a \ge b \ge c \ge 0$, that would make all three $0$, and there would be no quadratic. So $a > 0$.
4. Multiplying $a$, $b$ and $c$ all by the same positive number does not change the root or either condition. So we may choose $a = 1$ to keep things simple. Then $b = \frac{1 + c}{2}$, and $b^2 = 4c$ becomes
$$\frac{(1 + c)^2}{4} = 4c \quad\Longrightarrow\quad 1 + 2c + c^2 = 16c \quad\Longrightarrow\quad c^2 - 14c + 1 = 0.$$
5. By the quadratic formula,
$$c = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2} = 7 \pm 4\sqrt{3}.$$
6. We need $c \le a = 1$. Since $7 + 4\sqrt{3}$ is about $13.9$, it is too big. So $c = 7 - 4\sqrt{3}$, which is about $0.07$.
7. Then $b = \frac{1 + c}{2} = \frac{8 - 4\sqrt{3}}{2} = 4 - 2\sqrt{3}$.
8. The root is
$$x = -\frac{b}{2a} = -\frac{4 - 2\sqrt{3}}{2} = -2 + \sqrt{3}.$$

As a sanity check, this is about $-0.27$: negative, as it should be when all the coefficients are positive.

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

Both conditions stay true when $a$, $b$, $c$ are all scaled by the same number, so only their ratios matter, and setting $a = 1$ leaves just one unknown. The condition $a \ge c$ is what picks the right sign in step 6. For a quadratic with one root, remember the root is simply $-\frac{b}{2a}$.

Work with the root $r$ directly. A quadratic with the single root $r$ and $a = 1$ is $(x - r)^2 = x^2 - 2rx + r^2$, so $b = -2r$ and $c = r^2$. The arithmetic sequence condition $2b = a + c$ becomes $-4r = 1 + r^2$, or $r^2 + 4r + 1 = 0$, so $r = -2 \pm \sqrt{3}$. We also need $b \le a$, that is $-2r \le 1$, so $r \ge -\frac{1}{2}$. Only $r = -2 + \sqrt{3} \approx -0.27$ passes.

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