2021 AMC 12A Problems/Problem 2
Contents
- 1 Problem
- 2 Solution 1 (Algebra)
- 3 Solution 2 (Algebra)
- 4 Solution 3 (Process of Elimination)
- 5 Solution 4 (Graphing)
- 6 Video Solution (Quick and Easy)
- 7 Video Solution by Aaron He
- 8 Video Solution by Hawk Math
- 9 Video Solution by OmegaLearn (Using Logic and Analyzing Answer Choices)
- 10 Video Solution by TheBeautyofMath
- 11 See also
Problem
Under what conditions is true, 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 (Algebra)
One can 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 (Algebra)
Complete the square of the left side by rewriting the radical to be From there it is evident for the square root of the left to be equal to the right, must be equal to zero. Also, we know that the equivalency of square root values only holds true for nonnegative values of , making the correct answer
~AnkitAmc
Solution 3 (Process of Elimination)
The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need which refutes and Next, picking refutes and picking refutes By POE (Process of Elimination), the answer is
~MRENTHUSIASM
Solution 4 (Graphing)
If we graph then we get the union of:
- positive -axis
- positive -axis
- origin
Therefore, the answer is
The graph of is shown below. ~MRENTHUSIASM (credit given to TheAMCHub)
Video Solution (Quick and Easy)
~Education, the Study of Everything
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.