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what is the following quotient? \\frac{2-\\sqrt{8}}{4+\\sqrt{12}} \\cir…

Question

what is the following quotient?

\frac{2-\sqrt{8}}{4+\sqrt{12}}

\circ \frac{\sqrt{3}-\sqrt{6}}{4}

\circ \frac{2+\sqrt{3}-2\sqrt{2}-\sqrt{6}}{4}

\circ 2-\sqrt{3}-2\sqrt{2}+\sqrt{6}

\circ \frac{-2-\sqrt{3}+2\sqrt{2}+\sqrt{6}}{2}

Explanation:

Response

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</pre_analysis>

<reasoning>

Simplify the radicals

Using the Radical Simplification knowledge point
\[

$$\begin{aligned} \sqrt{8} &= \sqrt{4 \cdot 2} = 2\sqrt{2} \\ \sqrt{12} &= \sqrt{4 \cdot 3} = 2\sqrt{3} \end{aligned}$$

\]

Rewrite the expression

Using the Algebraic Simplification knowledge point
\[
\frac{2 - \sqrt{8}}{4 + \sqrt{12}} = \frac{2 - 2\sqrt{2}}{4 + 2\sqrt{3}} = \frac{2(1 - \sqrt{2})}{2(2 + \sqrt{3})} = \frac{1 - \sqrt{2}}{2 + \sqrt{3}}
\]

Rationalize the denominator

Using the Rationalizing the Denominator knowledge point
\[

$$\begin{aligned} \frac{1 - \sqrt{2}}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} &= \frac{(1 - \sqrt{2})(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2} \\ &= \frac{2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}}{4 - 3} \\ &= 2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(\frac{\sqrt{3} - \sqrt{6}}{4}\)</mcq-option>
<mcq-option>(B) \(\frac{2 + \sqrt{3} - 2\sqrt{2} - \sqrt{6}}{4}\)</mcq-option>
<mcq-correct>(C) \(2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}\)</mcq-correct>
<mcq-option>(D) \(\frac{-2 - \sqrt{3} + 2\sqrt{2} + \sqrt{6}}{2}\)</mcq-option>
</answer>

<post_analysis>
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"subject": "Mathematics",
"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Algebra",
"Rationalizing the Denominator"
]
}
</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

Simplify the radicals

Using the Radical Simplification knowledge point
\[

$$\begin{aligned} \sqrt{8} &= \sqrt{4 \cdot 2} = 2\sqrt{2} \\ \sqrt{12} &= \sqrt{4 \cdot 3} = 2\sqrt{3} \end{aligned}$$

\]

Rewrite the expression

Using the Algebraic Simplification knowledge point
\[
\frac{2 - \sqrt{8}}{4 + \sqrt{12}} = \frac{2 - 2\sqrt{2}}{4 + 2\sqrt{3}} = \frac{2(1 - \sqrt{2})}{2(2 + \sqrt{3})} = \frac{1 - \sqrt{2}}{2 + \sqrt{3}}
\]

Rationalize the denominator

Using the Rationalizing the Denominator knowledge point
\[

$$\begin{aligned} \frac{1 - \sqrt{2}}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} &= \frac{(1 - \sqrt{2})(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2} \\ &= \frac{2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}}{4 - 3} \\ &= 2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(\frac{\sqrt{3} - \sqrt{6}}{4}\)</mcq-option>
<mcq-option>(B) \(\frac{2 + \sqrt{3} - 2\sqrt{2} - \sqrt{6}}{4}\)</mcq-option>
<mcq-correct>(C) \(2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}\)</mcq-correct>
<mcq-option>(D) \(\frac{-2 - \sqrt{3} + 2\sqrt{2} + \sqrt{6}}{2}\)</mcq-option>
</answer>

<post_analysis>
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]
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</post_analysis>