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

Solution 1

Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$. By Vieta's formula 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 $9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}$.

Solution 3

  • no math here*

There are 2 cases. Case 1 is that $(x-a)(x-b)=0$ and $(x-b)(x-c)=0$. Lets test that 1st. If $x-b=0$, the maximum value for $x$ and $b$ is $9$. Then $b=9$ and $x=9.$ The next highest values are $7$and $8$ so $a=8$ and $c=9$. Therefore, $\frac{18+8+7}{2}= \boxed{\textbf{(D)}~16.5}$.

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 10 Problems and Solutions

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