The given expression is:

$\frac{2\sqrt{2} + 3\sqrt{5}}{3 - \sqrt{2}}$

To simplify this, we rationalize the denominator by multiplying both numerator and denominator by the conjugate of the denominator, $3 + \sqrt{2}$. Here's the process:

Step 1: Multiply numerator and denominator by the conjugate

$\frac{2\sqrt{2} + 3\sqrt{5}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{(2\sqrt{2} + 3\sqrt{5})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}$

Step 2: Simplify the denominator

Using the difference of squares formula: $(3 - \sqrt{2})(3 + \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7$

So, the denominator becomes $7$.

Step 3: Expand the numerator

$(2\sqrt{2} + 3\sqrt{5})(3 + \sqrt{2}) = (2\sqrt{2} \cdot 3) + (2\sqrt{2} \cdot \sqrt{2}) + (3\sqrt{5} \cdot 3) + (3\sqrt{5} \cdot \sqrt{2})$

Simplify term by term:

• $2\sqrt{2} \cdot 3 = 6\sqrt{2}$
• $2\sqrt{2} \cdot \sqrt{2} = 2 \cdot 2 = 4$
• $3\sqrt{5} \cdot 3 = 9\sqrt{5}$
• $3\sqrt{5} \cdot \sqrt{2} = 3\sqrt{10}$

Adding these, the numerator becomes: $6\sqrt{2} + 4 + 9\sqrt{5} + 3\sqrt{10}$

Step 4: Write the simplified expression

$\frac{2\sqrt{2} + 3\sqrt{5}}{3 - \sqrt{2}} = \frac{6\sqrt{2} + 4 + 9\sqrt{5} + 3\sqrt{10}}{7}$

Thus, the final simplified expression is: $\frac{6\sqrt{2} + 4 + 9\sqrt{5} + 3\sqrt{10}}{7}$

Would you like to see this written in a different format or have further steps broken down? Let me know! 😊

Related Questions:

1. How do you rationalize the denominator of any radical expression?
2. Why does multiplying by the conjugate simplify the denominator?
3. Can this result be simplified further for specific values of $\sqrt{2}$ or $\sqrt{5}$?
4. How would you handle this if the numerator and denominator had different variables?
5. How does this process differ for cube roots or higher powers?

Tip:

When rationalizing denominators, always check if the numerator can also be simplified after the operation—it can make final results neater!