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:

Simplify the radicals

Using the Radical Simplification knowledge point

$$
LATEXBLOCK0
$$

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

$$
LATEXBLOCK1
$$

Answer:

• (A) \(\frac{\sqrt{3} - \sqrt{6}}{4}\)
• (B) \(\frac{2 + \sqrt{3} - 2\sqrt{2} - \sqrt{6}}{4}\)
• (C) \(2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}\) (Correct answer)
• (D) \(\frac{-2 - \sqrt{3} + 2\sqrt{2} + \sqrt{6}}{2}\)