Question

simplify. rationalize the denominator.
\frac{-sqrt{3}}{9 + sqrt{2}}

Explanation:

Step1: Multiply numerator and denominator by the conjugate of the denominator.

The conjugate of \(9 + \sqrt{2}\) is \(9 - \sqrt{2}\). So we multiply the fraction by \(\frac{9 - \sqrt{2}}{9 - \sqrt{2}}\):
\[
\frac{-\sqrt{3}}{9 + \sqrt{2}} \times \frac{9 - \sqrt{2}}{9 - \sqrt{2}} = \frac{-\sqrt{3}(9 - \sqrt{2})}{(9 + \sqrt{2})(9 - \sqrt{2})}
\]

Step2: Expand the denominator using the difference of squares formula \((a + b)(a - b)=a^2 - b^2\).

For the denominator, \(a = 9\) and \(b=\sqrt{2}\), so:
\[
(9 + \sqrt{2})(9 - \sqrt{2})=9^2 - (\sqrt{2})^2 = 81 - 2 = 79
\]

Step3: Expand the numerator using the distributive property.

\[
-\sqrt{3}(9 - \sqrt{2})=-9\sqrt{3}+\sqrt{3}\times\sqrt{2}=-9\sqrt{3}+\sqrt{6}
\]

Step4: Combine the results.

Putting the numerator and denominator together, we get:
\[
\frac{-9\sqrt{3}+\sqrt{6}}{79}=\frac{\sqrt{6}-9\sqrt{3}}{79}
\]

Answer:

\(\frac{\sqrt{6}-9\sqrt{3}}{79}\)