2021 AMC 12A Problems/Problem 2

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 $\sqrt{a^2+b^2}=a+b$ true, 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 (Algebra)

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

~Jhawk0224

Solution 2 (Algebra)

Complete the square of the left side by rewriting the radical to be $\sqrt{(a+b)^{2}-2ab}.$ From there it is evident for the square root of the left to be equal to the right, $ab$ must be equal to zero. Also, we know that the equivalency of square root values only holds true for positive values of $a+b$ , making the correct answer $\boxed{\textbf{(D)}}.$

~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 $a+b\ge 0,$ which refutes $\textbf{(B)}$ and $\textbf{(E)}.$ Next, picking $(a,b)=(0,0)$ refutes $\textbf{(A)},$ and picking $(a,b)=(1,2)$ refutes $\textbf{(C)}.$ By POE (Process of Elimination), the answer is $\boxed{\textbf{(D)}}.$

~MRENTHUSIASM

Solution 4 (Graphing)

If we graph $\sqrt{x^2+y^2}=x+y,$ then we get the union of:

positive $x$ -axis

positive $y$ -axis

origin

Therefore, the answer is $\boxed{\textbf{(D)}}.$

The graph of $\sqrt{x^2+y^2}=x+y$ is shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -5; int xMax = 5; int yMin = -5; int yMax = 5; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); draw((xMax,0)--(0,0)--(0,yMax),red+linewidth(1.5)); [/asy]$ ~MRENTHUSIASM (credit given to TheAMCHub)

Video Solution (Quick and Easy)

https://youtu.be/WCrLIqVR0kc

~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)

https://youtu.be/Yp2NfDk_D-g

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/rEWS75W0Q54?t=71

~IceMatrix

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

Page

Toolbox

Search