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

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==Solution==
 
==Solution==
Squaring both sides, <math>a^2+b^2=a^2+2ab+b^2</math>. This means that <math>ab=0</math>, however, we also must have <math>a+b\geq0</math> since the original equation's left side must be non-negative. Thus, the answer is \boxed{\textbf{(D)}}
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Squaring both sides, <math>a^2+b^2=a^2+2ab+b^2</math>. This means that <math>ab=0</math>, however, we also must have <math>a+b\geq0</math> since the original equation's left side must be non-negative. Thus, the answer is <math>\boxed{\textbf{(D)}}</math>
  
 
~JHawk0224
 
~JHawk0224

Revision as of 16:25, 11 February 2021

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

Squaring both sides, $a^2+b^2=a^2+2ab+b^2$. This means that $ab=0$, however, we also must have $a+b\geq0$ since the original equation's left side must be non-negative. Thus, the answer is $\boxed{\textbf{(D)}}$

~JHawk0224

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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