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

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*no math here*
 
*no math here*
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There are 2 cases. Case 1 is that <math>(x-a)(x-b)=0</math> and <math>(x-b)(x-c)=0</math>. Lets test that 1st. If <math>x-b=0</math>, the maximum value for <math>x</math> and <math>b</math> is 9. Then b=9 and x=9. The next highest values are 7 and 8 so a=8 and c=9. Therefore, <math>\frac{18+8+7}{2}= \boxed{\textbf{(D)}~16.5}</math>.
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There are 2 cases. Case 1 is that <math>(x-a)(x-b)=0</math> and <math>(x-b)(x-c)=0</math>. Lets test that 1st. If <math>x-b=0</math>, the maximum value for <math>x</math> and <math>b</math> is <math>9</math>. Then <math>b=9</math> and <math>x=9.</math> The next highest values are <math>7</math>and <math>8</math>so <math>a=8</math> and <math>c=9</math>. Therefore, <math>\frac{18+8+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 16:19, 28 January 2018

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 (Problems • Answer Key • Resources)
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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