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2021 AMC 10B Problem 2

Problem 2 of 25EasierAlgebra

What is the value of (3−23)2+(3+23)2? \begin{aligned} &\sqrt{\left(3-2\sqrt{3}\right)^2} \\ &{}+\sqrt{\left(3+2\sqrt{3}\right)^2}? \end{aligned}

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Solution

Because u2=∣u∣,\sqrt{u^2}=|u|, the expression equals ∣3−23∣+∣3+23∣.|3-2\sqrt3|+|3+2\sqrt3|. Since 23>3,2\sqrt3>3, this becomes (23−3)+(3+23)=43.(2\sqrt3-3)+(3+2\sqrt3)=4\sqrt3. Thus, the correct answer is D .
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Tagged: radical · absolute value

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