Difference between revisions of "2015 AMC 10B Problems/Problem 14"
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==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)\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>. |
==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 14:23, 31 December 2020
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 1
Expanding the equation and combining like terms results in . By Vieta's formula the sum of the roots is . 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 .
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.