The expression $\frac{2}{\sqrt{2} + \sqrt{3}}$ can be simplified by rationalizing the denominator. Here’s how to do it:

Step 1: Multiply both numerator and denominator by the conjugate of the denominator.

The conjugate of $\sqrt{2} + \sqrt{3}$ is $\sqrt{2} - \sqrt{3}$. So, multiply both the numerator and the denominator by $\sqrt{2} - \sqrt{3}$:

$\frac{2}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{2(\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})}$

Step 2: Simplify the denominator.

The denominator is now in the form $(a+b)(a-b)$, which simplifies as:

$(\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$

Step 3: Final expression.

Now simplify the entire expression:

$\frac{2(\sqrt{2} - \sqrt{3})}{-1} = -2(\sqrt{2} - \sqrt{3})$

So, the simplified expression is:

$-2(\sqrt{2} - \sqrt{3}) = -2\sqrt{2} + 2\sqrt{3}$

Thus, $\frac{2}{\sqrt{2} + \sqrt{3}} = -2\sqrt{2} + 2\sqrt{3}$.

Would you like further details on this, or do you have any questions?

5 Related Questions:

1. How do you rationalize the denominator for more complex expressions?
2. What is the geometric meaning of conjugates in radicals?
3. Can you explain more on how the difference of squares formula works?
4. How do you handle expressions with cube roots?
5. What happens when you rationalize a denominator with complex numbers?

Tip:

When rationalizing the denominator, always multiply by the conjugate of the denominator, which helps eliminate the square roots.