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)}} |
| + | |||
| + | ~JHawk0224 | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=1|num-a=3}} | {{AMC12 box|year=2021|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:25, 11 February 2021
Problem
Under what conditions is 
true, 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
Squaring both sides, 
. This means that , however, we also must have 
since the original equation's left side must be non-negative. Thus, the answer is \boxed{\textbf{(D)}}
~JHawk0224
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. 
