Difference between revisions of "2019 AMC 8 Problems/Problem 20"
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+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eXJnG96Xqp4&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=21 | ||
==See Also== | ==See Also== |
Revision as of 14:35, 23 April 2021
Contents
Problem 20
How many different real numbers satisfy the equation
Solution 1
We have that if and only if . If , then , giving 2 solutions. If , then , giving 2 more solutions. All four of these solutions work, so the answer is . Further, the equation is a quartic in , so by the Fundamental Theorem of Algebra, there can be at most four real solutions.
Solution 2
We can expand to get , so now our equation is . Subtracting from both sides gives us . Now, we can factor the left hand side to get . If and/or equals , then the whole left side will equal . Since the solutions can be both positive and negative, we have solutions: (we can find these solutions by setting and equal to and solving for ). So the answer is .
~UnstoppableGoddess
Solution 3
Associated Video - https://www.youtube.com/watch?v=Q5yfodutpsw
Solution 4
https://youtu.be/5BXh0JY4klM (Uses a difference of squares & factoring method, different from above solutions)
Solution 5 (Fast)
https://www.youtube.com/watch?v=44vrsk_CbF8&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=2 ~ MathEx
Video Solution
-Happytwin
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eXJnG96Xqp4&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=21
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.