Higher November 2021 Paper 1 Q8
8 Solve \(x^2 = 5x + 24\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| 8 and −3 | M1 | for rearranging to get \(x^2 - 5x - 24\ (= 0)\) or \(-x^2 + 5x + 24\ (= 0)\) |
| M1 | for \((x \pm 8)(x \pm 3)\) or \((x + a)(x + b)\) where \(ab = -24\) or \(a + b = -5\) or substitution into formula, condoning one sign error eg \((x =) \dfrac{--5 \pm \sqrt{(-5)^2 - 4 \times 1 \times -24}}{2 \times 1}\) | |
| A1 | for 8 and \(-3\) |
Additional guidance
Can be implied by \((x - 8)(x + 3)\) or \((-x + 8)(x + 3)\)