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

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This article was made early. Please ignore it until 2021 AMC 10A Problems are released.
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==Problem==
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What is the value of <math>\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}</math>?
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<math>\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</math>
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==Solution 1==
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Note that the square root of any number squared is always the absolute value of the squared number because the square root function will only return a nonnegative 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>.
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~bjc, abhinavg0627 and JackBocresion
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==Solution 2==
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Let <math>x = \sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}</math>, then <math>x^2 = (3-2\sqrt{3})^2+2\sqrt{(-3)^2}+(3+2\sqrt3)^2</math>. The <math>2\sqrt{(-3)^2}</math> term is there due to difference of squares when you simplify <math>2ab</math> from <math>(a + b)^2</math>. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel
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uwu
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==Video Solution==
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https://youtu.be/HHVdPTLQsLc
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~Math Python
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== Video Solution by OmegaLearn ==
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https://youtu.be/Df3AIGD78xM
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~pi_is_3.14
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==Video Solution==
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https://youtu.be/v71C6cFbErQ
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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https://youtu.be/gLahuINjRzU?t=154
 +
 
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~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/DvpN56Ob6Zw?t=1
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~Interstigation
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==Video Solution by Mathematical Dexterity (50 Seconds)==
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https://www.youtube.com/watch?v=ScZ5VK7QTpY
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==Video Solution==
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https://youtu.be/3GHD62FK0xY
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~Education, the Study of Everything
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==See Also==
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{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 20:22, 14 April 2024

Problem

What is the value of $\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}$?

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

Solution 1

Note that the square root of any number squared is always the absolute value of the squared number because the square root function will only return a nonnegative 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

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 when you simplify $2ab$ from $(a + b)^2$. Simplifying the expression gives us $x^2 = 48$, so $x=\boxed{\textbf{(D)} ~4\sqrt{3}}$ ~ shrungpatel 

uwu

Video Solution

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

Video Solution by OmegaLearn

https://youtu.be/Df3AIGD78xM

~pi_is_3.14

Video Solution

https://youtu.be/v71C6cFbErQ

~savannahsolver

Video Solution by TheBeautyofMath

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

~IceMatrix

Video Solution by Interstigation

https://youtu.be/DvpN56Ob6Zw?t=1

~Interstigation

Video Solution by Mathematical Dexterity (50 Seconds)

https://www.youtube.com/watch?v=ScZ5VK7QTpY

Video Solution

https://youtu.be/3GHD62FK0xY

~Education, the Study of Everything

See Also

2021 AMC 10B (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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