Question

What is the following quotient?

Answer 1

Answer: Choice C) $2-\sqrt{3}-2\sqrt{2}+\sqrt{6}$

====================================

Work Shown:

Part 1

$\frac{2-\sqrt{8}}{4+\sqrt{12}}\\\\\\\frac{(2-\sqrt{8})(4-\sqrt{12})}{(4+\sqrt{12})(4-\sqrt{12})}\\\\\\\frac{2(4-\sqrt{12})-\sqrt{8}(4-\sqrt{12})}{(4)^2-(\sqrt{12})^2}\\\\\\\frac{8-2\sqrt{12}-4\sqrt{8}+\sqrt{8}*\sqrt{12}}{16-12}\\\\\\\frac{8-2\sqrt{12}-4\sqrt{8}+\sqrt{8*12}}{4}$

Part 2

$\frac{8-2\sqrt{12}-4\sqrt{8}+\sqrt{96}}{4}\\\\\\\frac{8-2\sqrt{4*3}-4\sqrt{4*2}+\sqrt{16*6}}{4}\\\\\\\frac{8-2\sqrt{4}*\sqrt{3}-4\sqrt{4}*\sqrt{2}+\sqrt{16}*\sqrt{6}}{4}\\\\\\\frac{8-2*2*\sqrt{3}-4*2*\sqrt{2}+4*\sqrt{6}}{4}\\\\\\\frac{8-4*\sqrt{3}-8*\sqrt{2}+4*\sqrt{6}}{4}$

Part 3

$\frac{8-4*\sqrt{3}-8*\sqrt{2}+4*\sqrt{6}}{4}\\\\\\\frac{4*2-4*1*\sqrt{3}-4*2*\sqrt{2}+4*\sqrt{6}}{4}\\\\\\\frac{4(2-\sqrt{3}-2\sqrt{2}+\sqrt{6})}{4}\\\\\\2-\sqrt{3}-2\sqrt{2}+\sqrt{6}$

This shows the answer is choice C.

----------------------------------------------

Explanation:

There are a lot of steps, so I broke things up into 3 sections or parts. The idea I'm applying is if the denominator is of the form $a+\sqrt{b}$, then multiplying top and bottom by $a-\sqrt{b}$ will rationalize the denominator. This is due to the difference of squares rule (step 3 of part 1). The following step shows squaring a square root has the two cancel out.

From that point, I simplified each square root and factored to cancel out a pair of '4's.

To expand out $(2-\sqrt{8})(4-\sqrt{12})$, we can use either the FOIL method, the box method, or the distribution property. I used the distribution property.