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2021 AMC 10B Problems/Problem 2: Difference between revisions

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==Problem==
==Problem==
What is the value of <cmath>\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?</cmath>
What is the value of <math>\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2}</math>?


<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>
<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>


==Solution==
==Solution 1==
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>.
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>.  
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
~bjc, abhinavg0627 and JackBocresion
 
==Solution 2==
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
 
==Solution 3 (Memorizing Roots)==
 
Memorizing your square roots from 1 - 10 are really important for cheesing some AMC problems, so try to memorize them.
 
\( \sqrt{3} \) is about 1.7.
 
We then substitute \( \sqrt{3} \) for 1.7 to solve this.
 
We get
 
<cmath>
\sqrt{(3-2 \cdot 1.7)^2} + \sqrt{(3+2 \cdot 1.7)^2}
</cmath>
 
<cmath>
= \sqrt{(-0.4)^2} + \sqrt{(6.4)^2}
</cmath>
 
<cmath>
= \sqrt{0.16} + \sqrt{6.4^2}
</cmath>
 
<cmath>
= 0.4 + 6.4
</cmath>
 
<cmath>
= 6.8
</cmath>
 
Looking at the answer choices, we see that <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math> gives <math>6.8</math> for when \( \sqrt{3} = 1.7 \).
 
~Pinotation


==Video Solution==
==Video Solution==
https://youtu.be/HHVdPTLQsLc
https://youtu.be/HHVdPTLQsLc
~Math Python
~Math Python
==Video Solution (EASY TO UNDERSTAND📈)==
https://www.youtube.com/watch?v=A1Li_jciTZY
~CalculaCore


== Video Solution by OmegaLearn ==
== Video Solution by OmegaLearn ==
https://youtu.be/Df3AIGD78xM
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==
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}
{{MAA Notice}}

Latest revision as of 13:55, 5 September 2025

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

Solution 3 (Memorizing Roots)

Memorizing your square roots from 1 - 10 are really important for cheesing some AMC problems, so try to memorize them.

\( \sqrt{3} \) is about 1.7.

We then substitute \( \sqrt{3} \) for 1.7 to solve this.

We get

\[\sqrt{(3-2 \cdot 1.7)^2} + \sqrt{(3+2 \cdot 1.7)^2}\]

\[= \sqrt{(-0.4)^2} + \sqrt{(6.4)^2}\]

\[= \sqrt{0.16} + \sqrt{6.4^2}\]

\[= 0.4 + 6.4\]

\[= 6.8\]

Looking at the answer choices, we see that $\boxed{\textbf{(D)} ~4\sqrt{3}}$ gives $6.8$ for when \( \sqrt{3} = 1.7 \).

~Pinotation

Video Solution

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

Video Solution (EASY TO UNDERSTAND📈)

https://www.youtube.com/watch?v=A1Li_jciTZY

~CalculaCore

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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