Rationalize the denominator.

17/3 + squareroot 11

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

1 answer

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 \( 3 - \sqrt{11} \):

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

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

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

  1. Next, simplify the numerator:

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

  1. 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:

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