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

Latest revision as of 19:46, 25 August 2023

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 $(x-b)(2x-(a+c))=0 \Rightarrow (x-b)\left(x-\frac{a+c}{2}\right)=0$ . 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 $9+\frac{8+7}{2} = 9+7.5 = \boxed{\textbf{(D)}~16.5}$ .

Video Solution 1

https://youtu.be/rkusS2sqCEo

~Education, the Study of Everything=

Video Solution 2

https://youtu.be/sgbVftpqBs0

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 4: / Line 4: @@
 <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==
-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>.
+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==
-Factoring out <math>(x-b)</math> from the equation yields <math>(x-b)(2x-(a+c))</math>
+Factoring out <math>(x-b)</math> from the equation yields <math>(x-b)(2x-(a+c))=0 \Rightarrow (x-b)\left(x-\frac{a+c}{2}\right)=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>.
+==Video Solution 1==
+https://youtu.be/rkusS2sqCEo
+~Education, the Study of Everything=
+==Video Solution 2==
+https://youtu.be/sgbVftpqBs0
+~savannahsolver
 ==See Also==
 {{AMC10 box|year=2015|ab=B|num-b=13|num-a=15}}
 {{MAA Notice}}

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