Difference between revisions of "2015 AMC 10B Problems/Problem 14"
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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== |
| + | |||
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>. | ||
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| + | ==Solution 2== | ||
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| + | 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== | ==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}} | ||
Latest revision as of 19:46, 25 August 2023
Problem
Let 
, 
, and 
be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation 
?
Solution 1
Expanding the equation and combining like terms results in 
. By Vieta's formula the sum of the roots is ![$\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}$](http://latex.artofproblemsolving.com/3/7/4/3743e38f1afcc5182d2a9da5e198ff8df18800ec.png)
. To maximize this expression we want to be the largest, and from there we can assign the next highest values to and . So let 
, 
, and 
. Then the answer is 
.
Solution 2
Factoring out 
from the equation yields 
. Therefore the roots are and 
. Because must be the larger root to maximize the sum of the roots, letting 
and be 
and 
respectively yields the sum 
.
Video Solution 1
~Education, the Study of Everything=
Video Solution 2
~savannahsolver
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
