Question

Rationalize the denominator.

17/3 + squareroot 11

17/3+ squareroot 11 = blank (Simplify your answer. Please type an exact answer, using radicals as needed.)

Answer 1

To rationalize the denominator of the expression \( \frac{17}{3 + \sqrt{11}} \), we will multiply the numerator and the denominator by the conjugate of the denominator, which is \( 3 - \sqrt{11} \).

Here's how this is done step by step:

1. Multiply the numerator and the denominator by \( 3 - \sqrt{11} \):

\[
\frac{17}{3 + \sqrt{11}} \cdot \frac{3 - \sqrt{11}}{3 - \sqrt{11}} = \frac{17(3 - \sqrt{11})}{(3 + \sqrt{11})(3 - \sqrt{11})}
\]

2. Now, we need to simplify the denominator using the difference of squares formula:

\[
(3 + \sqrt{11})(3 - \sqrt{11}) = 3^2 - (\sqrt{11})^2 = 9 - 11 = -2
\]

3. Next, simplify the numerator:

\[
17(3 - \sqrt{11}) = 51 - 17\sqrt{11}
\]

4. Putting it all together, we have:

\[
\frac{51 - 17\sqrt{11}}{-2} = -\frac{51}{2} + \frac{17\sqrt{11}}{2}
\]

Thus, the rationalized expression is:

\[
-\frac{51}{2} + \frac{17\sqrt{11}}{2}
\]

As a final answer, we can express this as:

\[
\frac{17\sqrt{11}}{2} - \frac{51}{2}
\]

So the complete simplified answer is:

\[
\frac{17\sqrt{11} - 51}{2}
\]