2015 AMC 12B Problems/Problem 12

Problem

Let $a$ , $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$ ?

$\textbf{(A)}\; 15 \qquad\textbf{(B)}\; 15.5 \qquad\textbf{(C)}\; 16 \qquad\textbf{(D)} 16.5 \qquad\textbf{(E)}\; 17$

Solution

Expand the polynomial. We get $(x-a)(x-b)+(x-b)(x-c)=x^2-(a+b)x+ab+x^2-(b+c)+bc=2x^2-(a+2b+c)x+(ab+bc).$

Now, consider a general quadratic equation $ax^2+bx+c=0.$ The two solutions to this are $\dfrac{-b}{2a}+\dfrac{\sqrt{b^2-4ac}}{2a}, \dfrac{-b}{2a}-\dfrac{\sqrt{b^2-4ac}}{2a}.$ The sum of these roots is $\dfrac{-b}{a}.$

Therefore, reconsidering the polynomial of the problem, the sum of the roots is $\dfrac{a+2b+c}{2}.$ Now, to maximize this, it is clear that $b=9.$ Also, we must have $a=8, b=7$ (or vice versa). The reason $a,b$ have to equal these values instead of larger values is because each of $a,b,c$ is distinct.

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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2015 AMC 12B Problems/Problem 12

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