We are asked to simplify the expression $\frac{6\sqrt{3} - 4}{8 - \sqrt{3}}$ and express the result in the form $p\sqrt{3} - q$, where $p$ and $q$ are positive rational numbers.

Step-by-Step Solution:

1. Multiply by the conjugate of the denominator to rationalize it. The conjugate of $8 - \sqrt{3}$ is $8 + \sqrt{3}$. So, multiply both the numerator and denominator by $8 + \sqrt{3}$:

$\frac{(6\sqrt{3} - 4)(8 + \sqrt{3})}{(8 - \sqrt{3})(8 + \sqrt{3})}$

2. Simplify the denominator using the difference of squares formula: $(8 - \sqrt{3})(8 + \sqrt{3}) = 8^2 - (\sqrt{3})^2 = 64 - 3 = 61$

Thus, the denominator is now 61.

3. Expand the numerator by distributing $6\sqrt{3} - 4$ over $8 + \sqrt{3}$: $(6\sqrt{3} - 4)(8 + \sqrt{3}) = 6\sqrt{3} \cdot 8 + 6\sqrt{3} \cdot \sqrt{3} - 4 \cdot 8 - 4 \cdot \sqrt{3}$

This simplifies as follows: $= 48\sqrt{3} + 6 \cdot 3 - 32 - 4\sqrt{3}$ $= 48\sqrt{3} + 18 - 32 - 4\sqrt{3}$ $= 44\sqrt{3} - 14$

4. Put it all together: $\frac{44\sqrt{3} - 14}{61}$

This can be written as: $\frac{44\sqrt{3}}{61} - \frac{14}{61}$

So, the final expression is: $\frac{44}{61}\sqrt{3} - \frac{14}{61}$

Thus, in the form $p\sqrt{3} - q$, we have $p = \frac{44}{61}$ and $q = \frac{14}{61}$.

Would you like further details on any part of the solution?

Related Questions:

1. How do you determine the conjugate of a binomial expression involving square roots?
2. Why do we use the difference of squares to rationalize the denominator?
3. Can you explain the importance of expanding and simplifying the numerator?
4. What other forms can this answer take if we change the requirement of $p$ and $q$ being rational numbers?
5. How can this problem be generalized to other square roots?

Tip:

Always remember to rationalize denominators when working with square roots, as it simplifies expressions and makes them easier to interpret.