Higher November 2022 Paper 2 Q19
19 Solve \(6x^2 + 5x - 6 = 0\) (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-\dfrac{3}{2}\) and \(\dfrac{2}{3}\) | M1 | for \((2x \pm 3)(3x \pm 2)\) or \((6x \pm 4)\left(x \pm \dfrac{9}{6}\right)\) or \((6x \pm 4)\left(x \pm \dfrac{3}{2}\right)\) or correct substitution into the quadratic formula, eg \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 6 \times (-6)}}{2 \times 6}\) |
| M1 | \((2x + 3)(3x - 2)\) \((6x - 4)\left(x + \dfrac{9}{6}\right)\) or \((6x - 4)\left(x + \dfrac{3}{2}\right)\) or \(\dfrac{-5 \pm \sqrt{169}}{12}\) or one correct answer | |
| A1 | oe accept answers of \(-1.5\) and in the range 0.66 to 0.67 |