what is the simplest form of $\frac{2sqrt{2}}{sqrt{3}-sqrt{2}}$?\n$2sqrt{6}+4$\n$2sqrt{5}+4$\n$\frac{2sqrt{6}…

what is the simplest form of $\frac{2sqrt{2}}{sqrt{3}-sqrt{2}}$?\n$2sqrt{6}+4$\n$2sqrt{5}+4$\n$\frac{2sqrt{6}+4}{5}$\n$\frac{2sqrt{5}+4}{5}$\ndone

what is the simplest form of $\frac{2sqrt{2}}{sqrt{3}-sqrt{2}}$?\n$2sqrt{6}+4$\n$2sqrt{5}+4$\n$\frac{2sqrt{6}+4}{5}$\n$\frac{2sqrt{5}+4}{5}$\ndone

Answer

Explanation:

Step1: Rationalize the denominator

Multiply the fraction $\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ by $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$. We get $\frac{2\sqrt{2}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$.

Step2: Expand the numerator and denominator

The numerator $2\sqrt{2}(\sqrt{3}+\sqrt{2})=2\sqrt{6}+4$. The denominator $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = (\sqrt{3})^2-(\sqrt{2})^2=3 - 2=1$. So the fraction is $\frac{2\sqrt{6}+4}{1}=2\sqrt{6}+4$.

Answer:

$2\sqrt{6}+4$