2015 AMC 10B Problems/Problem 14

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

Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$ . By Vieta's formulae the sum of the roots is $\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}$ . To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$ . So let $b=9$ , $a=8$ , and $c=7$ . Then the answer is $\dfrac{8+18+7}{2}=\boxed{\textbf{(D)} 16.5}$ .

Solution 2

Factoring out $(x-b)$ from the equation yields $(x-b)(2x-(a+c))=0 \Rightarrow (x-b)(x-\frac{a+c}{2})=0$ . Therefore the roots are $b$ and $\frac{a+c}{2}$ . Because $b$ must be the larger root to maximize the sum of the roots, letting $a,b,$ and $c$ be $8,9,$ and $7$ respectively yields the sum $9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}$ .

2015 AMC 10B (Problems • Answer Key • Resources)
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2015 AMC 10B Problems/Problem 14

Contents

Problem

Solution

Solution 2

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