Difference between revisions of "2015 AMC 10B Problems/Problem 14"

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==Solution==
 
==Solution==
Expanding the equation and combining like terms results in <math>2x^2-(a+2b+c)x+(ab+bc)=0</math>. By viettas the sum of the roots is <math>\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}</math>. To maximize this expression we want <math>b</math> to be the largest, and from there we can assign the next highest values to <math>a</math> and <math>c</math>. So let <math>b=9</math>, <math>a=8</math>, and <math>c=7</math>. Then the answer is <math>\dfrac{8+18+7}{2}=\boxed{\textbf{(D)} 16.5}</math>.
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Expanding the equation and combining like terms results in <math>2x^2-(a+2b+c)x+(ab+bc)=0</math>. By Vieta's formulae the sum of the roots is <math>\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}</math>. To maximize this expression we want <math>b</math> to be the largest, and from there we can assign the next highest values to <math>a</math> and <math>c</math>. So let <math>b=9</math>, <math>a=8</math>, and <math>c=7</math>. Then the answer is <math>\dfrac{8+18+7}{2}=\boxed{\textbf{(D)} 16.5}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2015|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:57, 6 March 2015

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 $\dfrac{8+18+7}{2}=\boxed{\textbf{(D)} 16.5}$.

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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