Question

What is the following quotient?

Answer 1

Answer: Choice C)
`2-√(3)-2√(2)+√(6)\\\\\\`

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Work Shown:

Part 1

`(2-√(8))/(4+√(12))\\\\\\((2-√(8))(4-√(12)))/((4+√(12))(4-√(12)))\\\\\\(2(4-√(12))-√(8)(4-√(12)))/((4)^2-(√(12))^2)\\\\\\(8-2√(12)-4√(8)+√(8)*√(12))/(16-12)\\\\\\(8-2√(12)-4√(8)+√(8*12))/(4)\\\\\\`

Part 2

`(8-2√(12)-4√(8)+√(96))/(4)\\\\\\(8-2√(4*3)-4√(4*2)+√(16*6))/(4)\\\\\\(8-2√(4)*√(3)-4√(4)*√(2)+√(16)*√(6))/(4)\\\\\\(8-2*2*√(3)-4*2*√(2)+4*√(6))/(4)\\\\\\(8-4*√(3)-8*√(2)+4*√(6))/(4)\\\\\\`

Part 3

`(8-4*√(3)-8*√(2)+4*√(6))/(4)\\\\\\(4*2-4*1*√(3)-4*2*√(2)+4*√(6))/(4)\\\\\\(4(2-√(3)-2√(2)+√(6)))/(4)\\\\\\2-√(3)-2√(2)+√(6)\\\\\\`

This shows the answer is choice C.

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Step-by-step 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+√(b)`, then multiplying top and bottom by
`a-√(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-√(8))(4-√(12))`, we can use either the FOIL method, the box method, or the distribution property. I used the distribution property.