Simplify. Rationalize the denominator. $$ \frac{-2}{6+\sqrt{2}} $$ $$ \sqrt{\square} $$

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We start with the expression

$$
\frac{-2}{6+\sqrt{2}}.
$$

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $$6-\sqrt{2}$$:

$$
\frac{-2}{6+\sqrt{2}} \cdot \frac{6-\sqrt{2}}{6-\sqrt{2}}.
$$

Multiplying the numerators:

$$
-2\left(6-\sqrt{2}\right) = -12+2\sqrt{2},
$$

and the denominators, using the difference of squares formula:

$$
(6+\sqrt{2})(6-\sqrt{2}) = 6^2 - (\sqrt{2})^2 = 36-2 = 34.
$$

Thus, the expression becomes:

$$
\frac{-12+2\sqrt{2}}{34}.
$$

We can simplify by factoring a 2 out of the numerator:

$$
\frac{2\left(-6+\sqrt{2}\right)}{34}.
$$

Now, cancel a factor of 2 from the numerator and denominator:

$$
\frac{-6+\sqrt{2}}{17}.
$$

Finally, we can write the answer in the preferred form:

$$
\frac{-6+\sqrt{2}}{17}.
$$

There is also a second line in the problem:

$$
\sqrt{\square}.
$$

If the box is intended as a placeholder for a square number to be placed under the square root symbol, note that our answer does not include an extra square root. The result is fully simplified and rationalized as above.
$$ \frac{-6 + \sqrt{2}}{17} $$

Simplify. Rationalize the denominator.

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