Higher June 2025 Paper 2R Q16
16 Show that \(\dfrac{4}{3\sqrt{5} + 7}\) can be written in the form \(a - \sqrt{b}\) where \(a\) and \(b\) are integers.
Show each stage of your working.
(3)
| Scheme | Marks |
|---|---|
| \(\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{3\sqrt{5} - 7}{3\sqrt{5} - 7}\) or \(\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{-3\sqrt{5} + 7}{-3\sqrt{5} + 7}\) oe | M1 |
eg \(\dfrac{4(3\sqrt{5} - 7)}{45 - 21\sqrt{5} + 21\sqrt{5} - 49}\) or \(\dfrac{4(3\sqrt{5} - 7)}{45 - 7^2}\) or \(\dfrac{12\sqrt{5} - 28}{45 - 21\sqrt{5} + 21\sqrt{5} - 49}\) or \(\dfrac{12\sqrt{5} - 28}{45 - 7^2}\) or \(\dfrac{4(3\sqrt{5} - 7)}{45 - 49}\) or \(\dfrac{4(3\sqrt{5} - 7)}{-4}\) or \(\dfrac{12\sqrt{5} - 28}{45 - 49}\) or \(\dfrac{12\sqrt{5} - 28}{-4}\) | M1 |
Working required Answer: \(7 - \sqrt{45}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for multiplying the numerator and denominator by \(3\sqrt{5} - 7\) or \(-3\sqrt{5} + 7\) (may be implied)
M1: for expanding the denominator in a correct fraction
denominator may be 4 terms which all need to be correct
\(\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{3\sqrt{5} - 7}{3\sqrt{5} - 7} = 7 - 3\sqrt{5}\) scores M1M0
Implies the 1st mark
A1: dep on M2
SCB1 for answer \(7 - \sqrt{45}\) with no method marks awarded
SCB2 for \(7 - \sqrt{45}\) if you would award the 1st M1 but not the 2nd M1 (total 2 marks)