2015 AMC 10B Problem 14

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 $\boxed{\textbf{(D)}\ 16.5}$.

A common factor should be pulled out, never multiplied through; it turns an unknown quadratic into two visible linear roots. Once the target is a weighted sum $b + \frac{a+c}{2}$, maximizing is a greedy assignment: the heaviest weight gets the biggest number. The distinctness condition is what stops the answer from being $9 + 9 = 18$.

Expand: $2x^2 - (a + 2b + c)x + (ab + bc) = 0$, so by Vieta the sum of the roots is $\frac{a + 2b + c}{2}$. Maximize $a + 2b + c$ with distinct digits: $8 + 18 + 7 = 33$, giving $16.5$.

• Expanding and applying Vieta carelessly, or maximizing by putting 9 in the wrong slot (the shared factor b counts fully, a and c only half).
• Ignoring "distinct" and using $a = b = c = 9$ to get $18$, which is not even a choice, then guessing $17$.