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

Revision as of 15:24, 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

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 (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

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 https://youtu.be/Df3AIGD78xM
+==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
 {{AMC10 box|year=2021|ab=B|num-b=1|num-a=3}}

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