Higher June 2025 Paper 3 Q22
22 Solve \(\quad 2x^2 \gt 12 - 5x\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2x^2 + 5x - 12\ (\gt 0)\) or \(12 - 5x - 2x^2\ (\lt 0)\) | B1 | oe may be implied by 1.5 and \(-4\) |
| Correct method to solve their three-term quadratic | M1 | eg \((2x - 3)(x + 4)\ (\gt 0)\) or \(2\left(x + \dfrac{5}{4}\right)^2 - \dfrac{121}{8}\ (\gt 0)\) or \(\dfrac{-5 \pm \sqrt{5^2 - 4 \times 2 \times -12}}{2 \times 2}\) any of these imply B1M1 |
| 1.5 and \(-4\) | A1ft | oe ft their three-term quadratic |
| Both \(x \gt 1.5\) and \(x \lt -4\) stated | B1ft | oe ft B0M1 or B1M1A0 ft their two solutions for their three-term quadratic correct for their inequality in the form \(ax^2 + bx + c \gt 0\) or \(ax^2 + bx + c \lt 0\), which must be seen if it is not the correct inequality |
Additional guidance
| For the first B1 and the M1 condone = 0 in place of \(\gt 0\) or \(\lt 0\) | |
| Trial and Improvement is 0, 3 (for 1.5 and \(-4\)) or 4 marks | |
| Incorrectly joined inequalities cannot score the final B1 eg \(1.5 \lt x \lt -4\) | B1M1A1B0 |
| Example where 1.5 and \(-4\) do not imply first B1: \(2x^2 + 5x - 12 \lt 0\), \((2x - 3)(x + 4) \lt 0\), \(x = 1.5\) and \(x = -4\), \(-4 \lt x \lt 1.5\) final B1ft is only awarded if B0M1 or B1M1A0 is scored | B0M1A1B1ft |
| \(2x^2 - 5x - 12 \lt 0\), \((2x + 3)(x - 4) \lt 0\), \(x = -1.5\) and \(x = 4\), \(-1.5 \lt x \lt 4\) joined inequality is correct for their three-term quadratic inequality | B0M1A1ftB1ft |