what is the following quotient?\n$\frac{2 - sqrt{8}}{4+sqrt{12}}$\n$\frac{sqrt{3}-sqrt{6}}{4}$\n$\frac{2+sqrt…

what is the following quotient?\n$\frac{2 - sqrt{8}}{4+sqrt{12}}$\n$\frac{sqrt{3}-sqrt{6}}{4}$\n$\frac{2+sqrt{3}-2sqrt{2}-sqrt{6}}{4}$\n$2-sqrt{3}-2sqrt{2}+sqrt{6}$\n$\frac{-2-sqrt{3}+2sqrt{2}+sqrt{6}}{2}$

what is the following quotient?\n$\frac{2 - sqrt{8}}{4+sqrt{12}}$\n$\frac{sqrt{3}-sqrt{6}}{4}$\n$\frac{2+sqrt{3}-2sqrt{2}-sqrt{6}}{4}$\n$2-sqrt{3}-2sqrt{2}+sqrt{6}$\n$\frac{-2-sqrt{3}+2sqrt{2}+sqrt{6}}{2}$

Answer

Explanation:

Step1: Simplify radicals

Simplify $\sqrt{8}=2\sqrt{2}$ and $\sqrt{12} = 2\sqrt{3}$. The expression becomes $\frac{2 - 2\sqrt{2}}{4+2\sqrt{3}}$.

Step2: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $4 - 2\sqrt{3}$. [ \begin{align*} &\frac{(2 - 2\sqrt{2})(4 - 2\sqrt{3})}{(4 + 2\sqrt{3})(4 - 2\sqrt{3})}\ =&\frac{2\times4-2\times2\sqrt{3}-2\sqrt{2}\times4 + 2\sqrt{2}\times2\sqrt{3}}{4^{2}-(2\sqrt{3})^{2}}\ =&\frac{8-4\sqrt{3}-8\sqrt{2}+4\sqrt{6}}{16 - 12}\ =&\frac{8-4\sqrt{3}-8\sqrt{2}+4\sqrt{6}}{4}\ =&\frac{8}{4}-\frac{4\sqrt{3}}{4}-\frac{8\sqrt{2}}{4}+\frac{4\sqrt{6}}{4}\ =&2-\sqrt{3}-2\sqrt{2}+\sqrt{6} \end{align*} ]

Answer:

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