Difference between revisions of "2018 AIME I Problems/Problem 5"
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The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>. | The cases <math>x=y</math> and <math>x=-2y</math> can be found by SFFT from <math>x^2+xy-2y^2=0 \implies (x+2y)(x-y)=0</math>. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=4|num-a=6}} | {{AIME box|year=2018|n=I|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:36, 14 May 2018
For each ordered pair of real numbers satisfying there is a real number such that Find the product of all possible values of .
Solution
Note that . That gives upon simplification and division by . Then, or . From the second equation, . If we take , we see that . If we take , we see that . The product is .
-expiLnCalc
Note
The cases and can be found by SFFT from .
See Also
2018 AIME I (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 | ||
All AIME Problems and Solutions |
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