Difference between revisions of "2015 AMC 10B Problems/Problem 14"
(→Solution 2) |
(→Solution 2) |
||
| Line 9: | Line 9: | ||
==Solution 2== | ==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))</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 10:51, 1 January 2016
Contents
Problem
Let 
, 
, and 
be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation 
?
Solution
Expanding the equation and combining like terms results in 
. By Vieta's formulae 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 
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. 
