Difference between revisions of "2018 AMC 12A Problems/Problem 5"
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==Solution== | ==Solution== | ||
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| + | We factor <math>x^2-3x+2</math> into <math>(x-1)(x-2)</math>. Thus, either <math>1</math> or <math>2</math> is a root of <math>x^2-5x+k</math>. If <math>1</math> is a root, then <math>1^2-5\cdot1+k=0</math>, so <math>k=4</math>. If <math>2</math> is a root, then <math>2^2-5\cdot2+k=0</math>, so <math>k=6</math>. The sum of all possible values of <math>k</math> is <math>\boxed{\textbf{(E)}10}</math>. | ||
==See Also== | ==See Also== | ||
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| − | + | {{AMC12 box|year=2018|ab=A|num-b=4|num-a=6}} | |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:25, 8 February 2018
Problem
What is the sum of all possible values of 
for which the polynomials 
and 
have a root in common?
Solution
We factor 
into 
. Thus, either 
or 
is a root of 
. If is a root, then 
, so 
. If is a root, then 
, so 
. The sum of all possible values of is 
.
See Also
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
