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Difference between revisions of "2021 AMC 12A Problems/Problem 2"

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==Solution 2 (Quick Inspection)==
 
==Solution 2 (Quick Inspection)==
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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{(B)}</math> and <math>\textbf{(E)}.</math> Next, picking <math>(a,b)=(1,2)</math> reveals that <math>\textbf{(C)}</math> is incorrect; picking <math>(a,b)=(0,0)</math> reveals that <math>\textbf{(A)}</math> is incorrect. By POE (Process of Elimination), the answer is <math>\boxed{\textbf{(D)}}.</math>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  

Revision as of 20:41, 12 February 2021

Problem

Under what conditions does $\sqrt{a^2+b^2}=a+b$ hold, where $a$ and $b$ are real numbers?

$\textbf{(A) }$ It is never true.

$\textbf{(B) }$ It is true if and only if $ab=0$.

$\textbf{(C) }$ It is true if and only if $a+b\ge 0$.

$\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$.

$\textbf{(E) }$ It is always true.

Solution 1

Square both sides to get $a^{2}+b^{2}=a^{2}+2ab+b^{2}$. Then, $0=2ab\rightarrow ab=0$. Also, it is clear that both sides of the equation must be nonnegative. The answer is $\boxed{\textbf{(D)}}$.

Solution 2 (Quick Inspection)

The left side of the original equation is the arithmetic square root, which is always nonnegative. So, we need $a+b\ge 0,$ which eliminates $\textbf{(B)}$ and $\textbf{(E)}.$ Next, picking $(a,b)=(1,2)$ reveals that $\textbf{(C)}$ is incorrect; picking $(a,b)=(0,0)$ reveals that $\textbf{(A)}$ is incorrect. By POE (Process of Elimination), the answer is $\boxed{\textbf{(D)}}.$ ~MRENTHUSIASM

Solution 3 (Graphing)

If we graph $\sqrt{x^2+y^2}=x+y,$ then we get the positive $x$-axis and the positive $y$-axis, plus the origin. Therefore, the answer is $\boxed{\textbf{(D)}}.$

Graph in Desmos: https://www.desmos.com/calculator/rlz6actjrw

~MRENTHUSIASM (credit given to TheAMCHub)

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn (Using logic and analyzing answer choices)

https://youtu.be/Yp2NfDk_D-g

~ 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. AMC logo.png

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