4400 Higher June 2008 Paper 3 Q19
19 Show that \(\dfrac{12}{\sqrt{8}} = 3\sqrt{2}\)
(2)
| Scheme | Marks |
|---|---|
eg \(\dfrac{12}{\sqrt{8}} = \dfrac{12}{2\sqrt{2}} = \dfrac{12}{2\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{12\sqrt{2}}{4}\) \(\dfrac{12}{\sqrt{8}} = \dfrac{12}{2\sqrt{2}} = \dfrac{6}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2}\) \(\dfrac{12}{\sqrt{8}} = \dfrac{12}{\sqrt{8}} \times \dfrac{\sqrt{8}}{\sqrt{8}} = \dfrac{12\sqrt{8}}{8} = \dfrac{3 \times 2\sqrt{2}}{2}\) \(\dfrac{12}{\sqrt{8}} = \dfrac{12}{\sqrt{8}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{12\sqrt{2}}{\sqrt{16}}\) | B1 B1 |
| (2) | |
| (2 marks) |
Notes
B1 B1: B1 for use of \(\sqrt{8} = 2\sqrt{2}\) or \(\sqrt{8} \times \sqrt{2} = \sqrt{16}\)
B1 for multiplication of numerator and denominator by \(\sqrt{2}\) or \(\sqrt{8}\) (in either order)
SC B1 for \(12 = 3\sqrt{16}\) or for both \(\left(\dfrac{12}{\sqrt{8}}\right)^2 = \dfrac{144}{8} = 18\) and \((3\sqrt{2})^2 = 9 \times 2 = 18\)
NB only total of 1 mark for either of these approaches