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

Line 16: Line 16:
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==
 
https://youtu.be/Df3AIGD78xM
 
https://youtu.be/Df3AIGD78xM
 +
 +
==Video Solution 3==
 +
https://youtu.be/v71C6cFbErQ
 +
 +
~savannahsolver
  
 
==Solution 2==
 
==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. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel
 
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. Simplifying the expression gives us <math>x^2 = 48</math>, so <math>x=\boxed{\textbf{(D)} ~4\sqrt{3}}</math> ~ shrungpatel
 
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}

Revision as of 20:39, 16 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 a squared number is the absolute value of the number because the square root function always gives a positive number. We know that $3-2\sqrt{3}$ is actually negative, thus the absolute value is not $3-2\sqrt{3}$ but $2\sqrt{3} - 3$. So the first term equals $2\sqrt{3}-3$ and the second term is $3+2\sqrt3$.

Summed up you get $\boxed{\textbf{(D)} ~4\sqrt{3}}$ ~bjc and abhinavg0627

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

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
Invalid username
Login to AoPS