Higher November 2017 Paper 1 Q21
21 Show that \(\dfrac{6 - \sqrt{8}}{\sqrt{2} - 1}\) can be written in the form \(a + b\sqrt{2}\) where \(a\) and \(b\) are integers. (3)
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(\dfrac{6-\sqrt{8}}{\sqrt{2}-1} \times \dfrac{\sqrt{2}+1}{\sqrt{2}+1}\) | \(2 + 4\sqrt{2}\) | M1 | for correct first step eg multiplies numerator and denominator by \(\sqrt{2} + 1\) condone missing brackets |
| \({= \dfrac{6\sqrt{2}+6-\sqrt{8}\sqrt{2}-\sqrt{8}}{2-1}}\) \({=6\sqrt{2}+6-4-2\sqrt{2}}\) | M1 | (dep) for expansion of numerator with 4 terms correct with or without signs or 3 out of exactly 4 terms correct | |
| A1 | for \(2 + 4\sqrt{2}\) oe or for stating \(a = 2\) and \(b = 4\) |