2015 AMC 10B Problem 14

1. Spot the common factor. Both products contain $(x - b)$. Pull it out:
$$(x - b)\big[(x - a) + (x - c)\big] = (x - b)(2x - a - c) = 0.$$
2. Read off the roots. A product is zero when one of its factors is zero.
- $x - b = 0$ gives $x = b$.
- $2x - a - c = 0$ gives $x = \dfrac{a + c}{2}$, the average of $a$ and $c$.
3. Write the sum. The sum of the roots is
$$b + \frac{a + c}{2}.$$
4. Make it as big as possible. The number $b$ counts in full, but $a$ and $c$ only count for half each. So the biggest digit should go into $b$: $b = 9$. The digits must all be different, so $a$ and $c$ get the next two biggest digits, $8$ and $7$ (in either order).
5. Compute.
$$9 + \frac{8 + 7}{2} = 9 + 7.5 = 16.5.$$

The answer is $\boxed{\textbf{(D)}\ 16.5}$.

When every term of an equation shares a factor, pull it out instead of multiplying everything out; the roots then appear one at a time. Once the sum is written as $b + \frac{a+c}{2}$, the biggest number belongs in the spot that counts the most. The rule that the digits are different is what stops the answer from being $9 + 9 = 18$.

Multiply everything out: $2x^2 - (a + 2b + c)x + (ab + bc) = 0$. For a quadratic $Ax^2 + Bx + C = 0$, the two roots add to $-\dfrac{B}{A}$, so here they add to $\dfrac{a + 2b + c}{2}$. Again $b$ has the most weight: $b = 9$, and $a, c = 8, 7$ give $\dfrac{8 + 18 + 7}{2} = 16.5$.

Both terms share the factor $(x - b)$, so
$$(x-b)\big[(x-a) + (x-c)\big] = (x-b)(2x - a - c) = 0.$$
The roots are $x = b$ and $x = \frac{a+c}{2}$, and their sum is
$$b + \frac{a + c}{2}.$$

The digit $b$ is counted at full weight while $a$ and $c$ each count for half, so put the largest available digit in $b$: $b = 9$. Then $a$ and $c$ must be the two largest remaining distinct digits, $8$ and $7$ (in either order). The maximum sum is
$$9 + \frac{8 + 7}{2} = 9 + 7.5 = 16.5.$$

The answer is $\boxed{\textbf{(D)}\ 16.5}$.