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

(Solution 3)
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<math>\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17</math>
 
<math>\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17</math>
  
==Solution==
+
==Solution 1==
  
===Solution 1===
 
 
Expanding the equation and combining like terms results in <math>2x^2-(a+2b+c)x+(ab+bc)=0</math>. By Vieta's formula 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>.
 
Expanding the equation and combining like terms results in <math>2x^2-(a+2b+c)x+(ab+bc)=0</math>. By Vieta's formula 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>.
  
===Solution 2===
+
==Solution 2==
  
 
Factoring out <math>(x-b)</math> from the equation yields <math>(x-b)(2x-(a+c))=0 \Rightarrow (x-b)(x-\frac{a+c}{2})=0</math>. Therefore the roots are <math>b</math> and <math>\frac{a+c}{2}</math>. Because <math>b</math> must be the larger root to maximize the sum of the roots, letting <math>a,b,</math> and <math>c</math> be <math>8,9,</math> and <math>7</math> respectively yields the sum <math>9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}</math>.
 
Factoring out <math>(x-b)</math> from the equation yields <math>(x-b)(2x-(a+c))=0 \Rightarrow (x-b)(x-\frac{a+c}{2})=0</math>. Therefore the roots are <math>b</math> and <math>\frac{a+c}{2}</math>. Because <math>b</math> must be the larger root to maximize the sum of the roots, letting <math>a,b,</math> and <math>c</math> be <math>8,9,</math> and <math>7</math> respectively yields the sum <math>9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}</math>.
 
===Solution 3===
 
 
*no math here*
 
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 17:47, 13 July 2020

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 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 $9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}$.

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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