Show that \( \frac{8 + \sqrt{12}}{5 + \sqrt{3}} \) can be written in the form \( \frac{a + \sqrt{3}}{b} \), where \( a \) and \( b \) are integers. - Edexcel - GCSE Maths - Question 20 - 2021 - Paper 1

Now expand the numerator:

$$(8 + \sqrt{12})(5 - \sqrt{3}) = 8 \cdot 5 - 8 \cdot \sqrt{3} + \sqrt{12} \cdot 5 - \sqrt{12} \cdot \sqrt{3}$$

Calculating each term, we have:

1. ( 8 \cdot 5 = 40 )
2. ( -8 \cdot \sqrt{3} )
3. ( \sqrt{12} \cdot 5 = 5\sqrt{12} )
4. ( \sqrt{12} \cdot \sqrt{3} = \sqrt{36} = 6 )

Putting it all together:

$$= 40 + 5\sqrt{12} - 6 - 8\sqrt{3} = 34 + 5\sqrt{12} - 8\sqrt{3}$$

Recognizing that ( \sqrt{12} = 2\sqrt{3} ), substituting gives us:

$$= 34 + 5(2\sqrt{3}) - 8\sqrt{3} = 34 + 10\sqrt{3} - 8\sqrt{3}$$

Thus combining like terms results in:

$$= 34 + 2\sqrt{3}$$

Now, we write the final expression as:

$$\frac{34 + 2\sqrt{3}}{22}$$

We can separate this into:

$$= \frac{34}{22} + \frac{2\sqrt{3}}{22}$$

Which simplifies to:

$$= \frac{17}{11} + \frac{\sqrt{3}}{11} \Rightarrow \frac{17 + \sqrt{3}}{11}$$

Thus, ( a = 17 ) and ( b = 11 ) are integers, thus proving the statement.