Difference between revisions of "2021 AMC 12A Problems/Problem 2"
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The left side of the original equation is the <b>arithmetic square root</b>, which is always nonnegative. So, we need <math>a+b\ge 0,</math> which eliminates <math>\textbf{(A)}</math> and <math>\textbf{(B)}.</math> Next, picking <math>(a,b)=(1,2)</math> reveals that <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math> are both incorrect. By POE (Process of Elimination), the answer is <math>\boxed{\textbf{(D)}}.</math> | The left side of the original equation is the <b>arithmetic square root</b>, which is always nonnegative. So, we need <math>a+b\ge 0,</math> which eliminates <math>\textbf{(A)}</math> and <math>\textbf{(B)}.</math> Next, picking <math>(a,b)=(1,2)</math> reveals that <math>\textbf{(C)}</math> and <math>\textbf{(E)}</math> are both incorrect. By POE (Process of Elimination), the answer is <math>\boxed{\textbf{(D)}}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 3 (Graphing)== | ||
+ | If we graph <math>\sqrt{x^2+y^2}=x+y,</math> then we get the positive <math>x</math>-axis and the positive <math>y</math>-axis, plus the origin. Therefore, the answer is <math>\boxed{\textbf{(D)}}.</math> | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Revision as of 04:43, 12 February 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
Square both sides to get . Then, . Then, the answer is . Consider a right triangle with legs and and hypotenuse . Then one of the legs must be equal to , but they are also nonnegative as they are lengths. Therefore, both and are correct.
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 and are both 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
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
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.