Higher June 2017 Paper 3 Q19
19 Solve \(2x^2 + 3x - 2 \gt 0\) (3)
| Answer | Mark | Notes |
|---|---|---|
| \(x \lt -2\), \(x \gt \dfrac{1}{2}\) | M1 | for a first step to solve the quadratic e.g. factorisation: \((2x + 4)\left(x - \tfrac{1}{2}\right)\) or \((2x - 1)(x + 2)\) or using the formula \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-2)}}{2 \times 2}\) |
| A1 | for \(-2\) and \(\tfrac{1}{2}\) | |
| A1 | for \(x \lt -2\), \(x \gt \tfrac{1}{2}\) |