Higher November 2022 Paper 2 Q19
19 Using the quadratic formula, or otherwise, solve \(\quad 3x^2 + x - 5 = 0\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{-1 \pm \sqrt{1^2 - 4 \times 3 \times -5}}{2 \times 3}\) or \(-\dfrac{1}{6} \pm \sqrt{\dfrac{5}{3} + \dfrac{1}{36}}\) | M1 | oe eg \(\dfrac{-1 \pm \sqrt{1 + 60}}{6}\) or \(-\dfrac{1}{6} \pm \sqrt{\dfrac{60}{36} + \dfrac{1}{36}}\) |
| \(\dfrac{-1 \pm \sqrt{61}}{6}\) or \(-\dfrac{1}{6} \pm \sqrt{\dfrac{61}{36}}\) or 1.135… and \(-1.468\)… | A1 | oe two solutions eg \(-\dfrac{1}{6} + \dfrac{1}{6}\sqrt{61}\) and \(-\dfrac{1}{6} - \dfrac{1}{6}\sqrt{61}\) allow decimal solutions rounded to at least 1 dp eg allow 1.14 and \(-1.5\) |
Additional guidance
| Both solutions correct | M1A1 |
| Both solutions seen in working but only one on answer line | M1A0 |
| Ignore conversion attempt after correct surd form solutions seen unless only one solution is subsequently selected | |
| Working must be for two solutions to score eg \(\dfrac{-1 + \sqrt{1^2 - 4 \times 3 \times -5}}{2 \times 3}\) not recovered | M0 |
| Square root sign should cover all appropriate work unless recovered eg \(-\dfrac{1}{6} \pm \sqrt{\dfrac{5}{3}} + \dfrac{1}{36}\) not recovered | M0 |
| Fraction line should be under all appropriate work unless recovered eg \(-1 \pm \dfrac{\sqrt{61}}{6}\) not recovered | M0 |
| One solution correct does not imply M1 | |
| Both solutions seen in working but signs transposed on answer line | M1A0 |
| √(12 – 4 × 3 × –5) is correct for \(\sqrt{1^2 - 4 \times 3 \times -5}\) |