Difference between revisions of "2021 AMC 10B Problems/Problem 2"

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==Solution==
 
==Solution==
Note that the square root of a squared number is the absolute value of the number. We know that <math>3-2\sqrt{3}</math> is actually negative, thus the absolute value is not <math>3-2\sqrt{3}</math> but <math>2\sqrt{3} - 3</math>.
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Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that <math>3-2\sqrt{3}</math> is actually negative, thus the absolute value is not <math>3-2\sqrt{3}</math> but <math>2\sqrt{3} - 3</math>.
 
So the first term equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>
 
So the first term equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>
 
Summed up you get <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~bjc and abhinavg0627
 
Summed up you get <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~bjc and abhinavg0627

Revision as of 19:36, 11 February 2021

Problem

What is the value of \[\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?\]

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3}+6$

Solution

Note that the square root of a squared number is the absolute value of the number because the square root function always gives a positive number. We know that $3-2\sqrt{3}$ is actually negative, thus the absolute value is not $3-2\sqrt{3}$ but $2\sqrt{3} - 3$. So the first term equals $2\sqrt{3}-3$ and the second term is $3+2\sqrt3$ Summed up you get $\boxed{\textbf{(D)} ~4\sqrt{3}}$ ~bjc and abhinavg0627

\phantom{no problem bjc}

Video Solution

https://youtu.be/HHVdPTLQsLc ~Math Python

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