Question

What is the simplest form 0f 2√2/√3-√2

Answer 1

Answer:

$\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}=2\sqrt{6}+4$

Step-by-step explanation:

We have the expression:

$\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

To simplify the expression multiply by the conjugate of the denominator.

The denominator is

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

Then its conjugate is:

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

Then:

$\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}*\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\\\=\frac{2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}\\\\\frac{2\sqrt{6}+4}{3-2}}\\\\= \frac{2\sqrt{6}+4}{1}}=2\sqrt{6}+4$

Answer 2

Answer:

$4+2\sqrt{6}$

Step-by-step explanation:

Given expression is $\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$.

Now we need to simplify this. So multiply and divide by conjugate of denominator.

$\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{2\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)}$

$=\frac{2\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)}\cdot\frac{\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)}$

$=\frac{2\sqrt{6}+2\sqrt{4}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}$

$=\frac{2\sqrt{6}+2\cdot2}{3-2}$

$=\frac{2\sqrt{6}+4}{1}$

$=2\sqrt{6}+4$

$=4+2\sqrt{6}$

Hence final answer is $4+2\sqrt{6}$.