Higher June 2019 Paper 2 Q16
16 Show that \(\dfrac{4 + \sqrt{8}}{\sqrt{2} - 1}\) can be written in the form \(a + b\sqrt{2}\), where \(a\) and \(b\) are integers.
Show each stage of your working clearly and give the value of \(a\) and the value of \(b\).
(3)
| Scheme | Marks |
|---|---|
| \(\dfrac{4 + \sqrt{8}}{\sqrt{2} - 1} \times \dfrac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)}\) | M1 |
e.g. \(\dfrac{4\sqrt{2} + 4 + \sqrt{8}\sqrt{2} + \sqrt{8}}{2 - 1}\) or \(\dfrac{4\sqrt{2} + 4 + 4 + \sqrt{8}}{2 - 1}\) or \(\dfrac{4\sqrt{2} + 4 + \sqrt{16} + \sqrt{8}}{2 - 1}\) or \(= 4\sqrt{2} + 4 + 4 + \sqrt{8}\) oe | M1 |
Working required Answer: \(8 + 6\sqrt{2}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for rationalising the denominator by multiplying numerator and denominator by \(\sqrt{2} + 1\) (or \(-\sqrt{2} - 1\))
condone missing brackets
M1: (dep) for expansion of numerator with at least 3 terms correct oe
Using \(-\sqrt{2} - 1\)
e.g.
\(\dfrac{-4\sqrt{2} - 4 - \sqrt{8}\sqrt{2} - \sqrt{8}}{-2 + 1}\) or \(\dfrac{-4\sqrt{2} - 4 - 4 - \sqrt{8}}{-2 + 1}\) or
\(\dfrac{-4\sqrt{2} - 4 - \sqrt{16} - \sqrt{8}}{-2 + 1}\)
A1: (dep on M2) or for stating \(a = 8\) and \(b = 6\)