Higher June 2024 Paper 1 Q17
17 Show that \(\;\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}}\;\) can be written in the form \(a + \sqrt{b}\;\) where \(a\) and \(b\) are integers.
Show each stage of your working clearly.
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3 + \sqrt{5}}{3 + \sqrt{5}}\) oe or \(\dfrac{1 + \sqrt{5}}{3 - \sqrt{5}} \times \dfrac{-3 - \sqrt{5}}{-3 - \sqrt{5}}\) oe | M1 |
\(\dfrac{3 + \sqrt{5} + 3\sqrt{5} + \sqrt{5}\sqrt{5}}{9 + 3\sqrt{5} - 3\sqrt{5} - \sqrt{5}\sqrt{5}}\) oe or \(\dfrac{3 + \sqrt{5} + 3\sqrt{5} + \sqrt{5}\sqrt{5}}{9 - \sqrt{5}\sqrt{5}}\) oe or \(\dfrac{3 + \sqrt{5} + 3\sqrt{5} + 5}{9 + 3\sqrt{5} - 3\sqrt{5} - 5}\) oe or \(\dfrac{8 + \sqrt{5} + 3\sqrt{5}}{9 - 5}\) oe or \(\dfrac{3 + 4\sqrt{5} + 5}{9 - 5}\) oe or \(\dfrac{8 + 4\sqrt{5}}{4}\) | M1 |
Working required Answer: \(2 + \sqrt{5}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for rationalising the denominator by multiplying numerator and denominator by \(3 + \sqrt{5}\) (or \(-3 - \sqrt{5}\))
M1: numerator correctly expanded and may be simplified to at least 2 terms and denominator correctly expanded and may be simplified to 1 term
A1: for \(2 + \sqrt{5}\) from correct working dep on M2