Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (Video solutions should be first come first serve. Also, made the title more uniform.)
m (→‎Solution 1)
Line 5: Line 5:
  
 
==Solution 1==
 
==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 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>.
  
 
~bjc, abhinavg0627 and JackBocresion
 
~bjc, abhinavg0627 and JackBocresion

Revision as of 15:37, 5 September 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 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. 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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_2&oldid=161638"