Higher June 2024 Paper 4 Q19
19 Show that \(\dfrac{\sqrt{3} + 2}{\sqrt{48} - 6}\) can be written in the form \(\dfrac{a + b\sqrt{3}}{6}\).
You must show each step in your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(\dfrac{\sqrt{3} + 2}{\sqrt{48} - 6} \times \dfrac{\sqrt{48} + 6}{\sqrt{48} + 6}\) | M1 | multiply by conjugate of denominator | |
| \(\dfrac{\sqrt{3} \times \sqrt{48} + 6\sqrt{3} + 2\sqrt{48} + 2 \times 6}{\sqrt{48} \times \sqrt{48} + 6\sqrt{48} - 6\sqrt{48} - 6 \times 6}\) | M1 | may be in a separate table FT \(\sqrt{48} - 6\) in both numerator and denominator | multiply out numerator giving at least three terms and denominator accept equivalents e.g. denominator as 48 – 36 or 12 |
| \(\sqrt{48} = 4\sqrt{3}\) soi | M1 | implied by [\(2\sqrt{48} =\)] \(8\sqrt{3}\) or \(14\sqrt{3}\) | |
| Simplifying their fraction e.g \(\frac{24 + 14\sqrt{3}}{48 - 36}\) or better | M1dep | Dep on at least three terms in the numerator FT their fraction with surds | e.g. collecting like terms in numerator and in the denominator |
| \(\dfrac{12 + 7\sqrt{3}}{6}\) | A1 | A1 dep on M4 | |