Question

What is the following quotient? 2- sqrt 8 / 4 + sqrt 12

Answer 1

C is the answer.

We multiply the top and bottom of this by the conjugate (switch the sign of the square root):

(2-sqrt8)(4-sqrt12)/(4+sqrt12)(4-sqrt12)

Multiying the top we have
8-2sqrt12-4sqrt8+sqrt(8×12)

Multiplying the bottom we have:
16-4sqrt12+4sqrt12-12

On top, we simplify the radicals and get
8-4sqrt3-8sqrt2+4sqrt6

On bottom, we have
16-12=4

Now we divide everything on top by 4 and get
2-sqrt3-2sqrt2+sqrt6

Answer 2

we have

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

Multiply the numerator and denominator by the conjugate of the denominator

so

$\frac{2-\sqrt{8}}{4+\sqrt{12}}*\frac{4-\sqrt{12}}{4-\sqrt{12}}=\frac{(2-\sqrt{8})(4-\sqrt{12)}}{4^{2}-(\sqrt{12})^{2}}\\ \\= \frac{2*4-2*\sqrt{12}-4*\sqrt{8}+\sqrt{12}*\sqrt{8}}{16-12}\\ \\=\frac{8-4\sqrt{3}-8\sqrt{2}+4\sqrt{6}}{4} \\ \\=2-\sqrt{3}-2\sqrt{2}+\sqrt{6}$

therefore

the answer is

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