Difference between revisions of "2021 AMC 12A Problems/Problem 2"
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==Solution 1== | ==Solution 1== | ||
− | Square both sides to get <math>a^{2}+b^{2}=a^{2}+2ab+b^{2}</math>. Then, <math>0=2ab\rightarrow ab=0</math>. Also, it is clear that both sides of the equation must be nonnegative. The answer is <math>\boxed{\textbf{(D)}}</math>. | + | Square both sides to get <math>a^{2}+b^{2}=a^{2}+2ab+b^{2}</math>. Then, <math>0=2ab\rightarrow ab=0</math>. Also, it is clear that both sides of the equation must be nonnegative. The answer is <math>\boxed{\textbf{(D)}}</math>. |
+ | |||
+ | ~Jhawk0224 | ||
==Solution 2 (Quick Inspection)== | ==Solution 2 (Quick Inspection)== |
Revision as of 17:34, 3 March 2021
Contents
[hide]Problem
Under what conditions does hold, where and are real numbers?
It is never true.
It is true if and only if .
It is true if and only if .
It is true if and only if and .
It is always true.
Solution 1
Square both sides to get . Then, . Also, it is clear that both sides of the equation must be nonnegative. The answer is .
~Jhawk0224
Solution 2 (Quick Inspection)
The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need which eliminates and Next, picking reveals that is incorrect, and picking reveals that is incorrect. By POE (Process of Elimination), the answer is
~MRENTHUSIASM
Solution 3 (Graphing)
If we graph then we get the positive -axis and the positive -axis, plus the origin. Therefore, the answer is
Graph in Desmos: https://www.desmos.com/calculator/0p3w7auwde
~MRENTHUSIASM (credit given to TheAMCHub)
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=40
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn (Using logic and analyzing answer choices)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/rEWS75W0Q54?t=71
~IceMatrix
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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.