4400 Higher June 2007 Paper 4 Q14
14 Solve \(\dfrac{5x - 7}{x - 1} = x + 1\)
(4)
| Scheme | Marks |
|---|---|
| \(5x - 7 = x^2 - 1\) or \(5x - 7 = (x - 1)(x + 1)\) | M1 |
| \(x^2 - 5x + 6 = 0\) | M1 |
| \((x - 2)(x - 3)\) (= 0) or \(\dfrac{5 \pm \sqrt{(-5)^2 - 4 \times 6}}{2}\) | M1 |
| \(x = 2\) or 3 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: condone \(5x - 7 = x - 1 \times x + 1\)
M1: allow different order with = 0
M1: \((x - 2.5)^2 + 6 - 6.25\)
A1: T & I or no wking: 4 mks or 0 mks