2021 AMC 10B Problems/Problem 2

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
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