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

m (→‎Solution 4 (Algebra))
(Rearranged solutions by titles.)
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~Jhawk0224
 
~Jhawk0224
  
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==Solution 2 (Process of Elimination)==
+
==Solution 2 (Algebra)==
 +
Complete the square of the left side by rewriting the radical to be <cmath>\sqrt{(x+y)^{2}-2ab}.</cmath>
 +
From there it is evident for the square root of the left to be equal to the right, <math>ab</math> must be equal to zero.
 +
Also we know that the equivalency of square root values only hold true for positive values of <math>a+b</math>, making the correct answer <math>\boxed{\textbf{(D)}}.</math>
 +
 
 +
~AnkitAmc
 +
 
 +
==Solution 3 (Process of Elimination)==
 
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 refutes <math>\textbf{(B)}</math> and <math>\textbf{(E)}.</math> Next, picking <math>(a,b)=(0,0)</math> refutes <math>\textbf{(A)},</math> and picking <math>(a,b)=(1,2)</math> refutes <math>\textbf{(C)}.</math> 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 refutes <math>\textbf{(B)}</math> and <math>\textbf{(E)}.</math> Next, picking <math>(a,b)=(0,0)</math> refutes <math>\textbf{(A)},</math> and picking <math>(a,b)=(1,2)</math> refutes <math>\textbf{(C)}.</math> By POE (Process of Elimination), the answer is <math>\boxed{\textbf{(D)}}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 3 (Graphing)==
+
==Solution 4 (Graphing)==
 
If we graph <math>\sqrt{x^2+y^2}=x+y,</math> then we get the union of:
 
If we graph <math>\sqrt{x^2+y^2}=x+y,</math> then we get the union of:
  
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</asy>
 
</asy>
 
~MRENTHUSIASM (credit given to TheAMCHub)
 
~MRENTHUSIASM (credit given to TheAMCHub)
−
 
−
==Solution 2 (Algebra)==
 
−
Complete the square of the left side by rewriting the radical to be <cmath>\sqrt{(x+y)^{2}-2ab}.</cmath>
 
−
From there it is evident for the square root of the left to be equal to the right, <math>ab</math> must be equal to zero.
 
−
Also we know that the equivalency of square root values only hold true for positive values of <math>a+b</math>, making the correct answer <math>\boxed{\textbf{(D)}}.</math>
 
−
 
−
~AnkitAmc
 
  
 
==Video Solution by Aaron He==
 
==Video Solution by Aaron He==

Revision as of 18:07, 12 July 2022

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)

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 \[\sqrt{(x+y)^{2}-2ab}.\] 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 hold 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 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

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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