Answer: Choice C) [tex]2-\sqrt{3}-2\sqrt{2}+\sqrt{6}\\\\\\[/tex]
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Work Shown:
Part 1
[tex]\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}\\\\\\[/tex]
Part 2
[tex]\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}\\\\\\[/tex]
Part 3
[tex]\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}\\\\\\[/tex]
This shows the answer is choice C.
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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 [tex]a+\sqrt{b}[/tex], then multiplying top and bottom by [tex]a-\sqrt{b}[/tex] 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 [tex](2-\sqrt{8})(4-\sqrt{12})[/tex], we can use either the FOIL method, the box method, or the distribution property. I used the distribution property.