16 (a) Rationalise the denominator of $$\frac{22}{\sqrt{11}}$$ Give your answer in its simplest form - Edexcel - GCSE Maths - Question 17 - 2019 - Paper 1

We start with the expression:

$\frac{\sqrt{3}}{2\sqrt{3} - 1}$

To simplify this, we multiply the numerator and the denominator by the conjugate of the denominator, which is $2\sqrt{3} + 1$:

$\frac{\sqrt{3}(2\sqrt{3} + 1)}{(2\sqrt{3} - 1)(2\sqrt{3} + 1)}$

Calculating the denominator:

$(2\sqrt{3})^2 - 1^2 = 12 - 1 = 11$

Now, simplifying the numerator:

$\sqrt{3}(2\sqrt{3}) + \sqrt{3}(1) = 2 \cdot 3 + \sqrt{3} = 6 + \sqrt{3}$

Thus, we have:

$\frac{6 + \sqrt{3}}{11}$

In this form, we see that $a = 6$ and $b = 11$, both of which are integers.