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

Revision as of 17:29, 11 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. 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

@@ Line 1: / Line 1: @@
-~~Problem~~
+==Problem==
-What is the value of <cmath>\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}?</cmath>
+What is the value of <cmath>\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}?</cmath>
 <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==
+Note that the square root of a squared number is the absolute value of the number.
-We have that <math>\sqrt{(3-2\sqrt{2})^2} = 3-2\sqrt{2}</math> and <math>\sqrt{(3+2\sqrt{2})^2} = 3+2\sqrt{2}</math>. Adding them together, we get that it is equal to <math>\boxed{6}</math>. ~ ArduinoRasp34567
+So the first term equals <math>2\sqrt{3}-3</math> and the second term is <math>3+2\sqrt3</math>
+Summed up you get <math>\boxed{\textbf{(D)} ~4\sqrt{3}}</math>~bjc

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

Revision as of 17:29, 11 February 2021

Problem

Solution