Question

rationalize the denominator.

\\\frac{6}{\sqrt{6} - \sqrt{3}}\\

\\\frac{6}{\sqrt{6} - \sqrt{3}} = \square\\

(simplify your answer. type an ...)

Explanation:

⚡ Using what you learned: Rationalizing Denominators

Step 1: Multiply by the conjugate

$$ \frac{6}{\sqrt{6} - \sqrt{3}} \cdot \frac{\sqrt{6} + \sqrt{3}}{\sqrt{6} + \sqrt{3}} = \frac{6(\sqrt{6} + \sqrt{3})}{(\sqrt{6})^2 - (\sqrt{3})^2} $$

Step 2: Simplify the denominator

$$ (\sqrt{6})^2 - (\sqrt{3})^2 = 6 - 3 = 3 $$

$$ \frac{6(\sqrt{6} + \sqrt{3})}{3} $$

Step 3: Divide out common factors

$$ 2(\sqrt{6} + \sqrt{3}) = 2\sqrt{6} + 2\sqrt{3} $$

Answer:

\(2\sqrt{6} + 2\sqrt{3}\) (or \(2(\sqrt{6} + \sqrt{3})\))