Show that \( \frac{6 - \sqrt{8}}{\sqrt{2} - 1} \) can be written in the form \( a + b\sqrt{2} \) where \( a \) and \( b \) are integers. - Edexcel - GCSE Maths - Question 21 - 2017 - Paper 1

The denominator simplifies as follows:

$$(\sqrt{2} - 1)(\sqrt{2} + 1) = \sqrt{2}^2 - 1^2 = 2 - 1 = 1$$

Next, let's expand the numerator:

$$(6 - \sqrt{8})(\sqrt{2} + 1) = 6\sqrt{2} + 6 - \sqrt{8}\sqrt{2} - \sqrt{8} = 6\sqrt{2} + 6 - 2\sqrt{2} - \sqrt{8} = (6 - 2)\sqrt{2} + 6 - 2\sqrt{2} = 4\sqrt{2} + 6 - 2\sqrt{2} = 4\sqrt{2} + 4$$

Thus, the expression simplifies to:

$$4 + 4\sqrt{2}$$

From the simplified expression, we can write:

$$4 + 4\sqrt{2} = a + b\sqrt{2}$$

where ( a = 4 ) and ( b = 4 ), which are both integers. Therefore, we have successfully shown that the original expression can be written in the required form.