Higher June 2023 Paper 1 Q24
24.
(a) Solve the equation \(5x^2 - 8x - 1 = 0\).
Give your answers in the form \(\dfrac{m \pm \sqrt{n}}{5}\), where \(m\) and \(n\) are integers. [3]
Give your answers in the form \(\dfrac{m \pm \sqrt{n}}{5}\), where \(m\) and \(n\) are integers. [3]
(b) Use factorisation to solve the following equation. [6]\[\frac{4}{x-1} = 3 + \frac{2}{x}\]
| Scheme | Marks | Comments |
|---|---|---|
| \((x =)\ \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4\times5\times(-1)}}{2\times5}\) | M1 | |
| \((x =)\ \dfrac{8 \pm \sqrt{84}}{10}\) | A1 | |
| \((x =)\ \dfrac{4 \pm \sqrt{21}}{5}\) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(4x = 3x(x - 1) + 2(x - 1)\) or \(4x = (3x + 2)(x - 1)\) oe, si | M2 | M1 for \(\left(\dfrac{4}{x-1}\right) = \dfrac{3x+2}{x}\) si |
| \(4x = 3x^2 + 2x - 3x - 2\) or better | M1 | Allow one error in expansion. |
| \(3x^2 - 5x - 2 = 0\) | A1 | CAO |
| \((3x + 1)(x - 2)\ (= 0)\) | M1 | Allow for \((3x - 1)(x + 2)\) FT ‘their \(3x^2 - 5x - 2 = 0\)’ provided at least M2 previously awarded. |
| \(x = -\dfrac{1}{3},\ x = 2\) | A1 |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| \(4 = \left(3 + \dfrac{2}{x}\right)(x - 1)\) | M1 | |
| \(4 = 3x - 3 + \dfrac{2x}{x} - \dfrac{2}{x}\) | M1 | Allow one error in expansion. |
| \(4x = 3x^2 - 3x + 2x - 2\) | M1 | |
| \(3x^2 - 5x - 2 = 0\) | A1 | CAO |
| \((3x + 1)(x - 2)\ (= 0)\) | M1 | Allow for \((3x - 1)(x + 2)\) FT ‘their \(3x^2 - 5x - 2 = 0\)’ provided at least M2 previously awarded. |
| \(x = -\dfrac{1}{3},\ x = 2\) | A1 |
| (9) |