Art of Problem Solving

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

Mc21s (talk | contribs)
No edit summary
MRENTHUSIASM (talk | contribs)
m Video solutions should be first come first serve. Also, made the title more uniform.
Line 4: Line 4:
<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 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>.
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>.


Line 15: Line 15:
https://youtu.be/HHVdPTLQsLc
https://youtu.be/HHVdPTLQsLc
~Math Python
~Math Python
==Video Solution (50 seconds) by Mathematical Dexterity==
https://www.youtube.com/watch?v=ScZ5VK7QTpY


== Video Solution by OmegaLearn ==
== Video Solution by OmegaLearn ==
https://youtu.be/Df3AIGD78xM
https://youtu.be/Df3AIGD78xM


==Video Solution 4==
==Video Solution==
https://youtu.be/v71C6cFbErQ
https://youtu.be/v71C6cFbErQ


Line 36: Line 33:


~Interstigation
~Interstigation
==Video Solution by Mathematical Dexterity (50 Seconds)==
https://www.youtube.com/watch?v=ScZ5VK7QTpY


==See Also==
==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}}
{{MAA Notice}}

Revision as of 20:08, 30 June 2021

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 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

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

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

Video Solution by OmegaLearn

https://youtu.be/Df3AIGD78xM

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=101

~Interstigation

Video Solution by Mathematical Dexterity (50 Seconds)

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

See Also

2021 AMC 10B (ProblemsAnswer KeyResources)
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 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.