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 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>.
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Note that the square root of any square is always the absolute value of the squared number because the square root function will only return a positive number. By squaring both <math>3</math> and <math>2\sqrt{3}</math>, we see that <math>2\sqrt{3}>3</math>, thus <math>3-2\sqrt{3}</math> is negative, so we must take the absolute value of <math>3-2\sqrt{3}</math>, which is just <math>2\sqrt{3}-3</math>. Knowing this, the first term in the expression equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>, and summing the two gives <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</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
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~bjc, abhinavg0627 and JackBocresion
  
 
==Video Solution==
 
==Video Solution==

Revision as of 15:12, 19 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 any square is always the absolute value of the squared number because the square root function will only return a positive number. By squaring both $3$ and $2\sqrt{3}$, we see that $2\sqrt{3}>3$, thus $3-2\sqrt{3}$ is negative, so we must take the absolute value of $3-2\sqrt{3}$, which is just $2\sqrt{3}-3$. Knowing this, the first term in the expression equals $2\sqrt{3}-3$ and the second term is $3+2\sqrt3$, and summing the two gives $\boxed{\textbf{(D)} ~4\sqrt{3}}$.

~bjc, abhinavg0627 and JackBocresion

Video Solution

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

Video Solution by OmegaLearn

https://youtu.be/Df3AIGD78xM

Video Solution 3

https://youtu.be/v71C6cFbErQ

~savannahsolver

Solution 2

Let $x = \sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}$, then $x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2$. The $2\sqrt{(-3)^2}$ term is there due to difference of squares. Simplifying the expression gives us $x^2 = 48$, so $x=\boxed{\textbf{(D)} ~4\sqrt{3}}$ ~ shrungpatel

Video Solution by TheBeautyofMath

https://youtu.be/gLahuINjRzU?t=154

~IceMatrix

2021 AMC 10B (ProblemsAnswer KeyResources)
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Problem 1
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Problem 3
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All AMC 10 Problems and Solutions
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