Higher November 2024 Paper 2 Q16
16 Solve \(\quad (x + 2)(x - 5) = 6x\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 + 2x - 5x - 10\) or \(x^2 - 3x - 10\) | M1 | oe quadratic expression 4 terms with at least 3 correct (terms may be seen in a grid) implied by \(x^2 - 3x + k\) |
| their \((x^2 + 2x - 5x - 10) - 6x\ (= 0)\) or their \((x^2 - 3x - 10) - 6x\ (= 0)\) or \(x^2 - 9x - 10\ (= 0)\) | M1dep | accept eg \(-x^2 + 9x + 10\ (= 0)\) accept oe equations in the form \(px^2 + qx = r\) eg \(x^2 - 9x = 10\) |
| \((x + 1)(x - 10)\ (= 0)\) or \(\dfrac{--9 \pm \sqrt{(-9)^2 - 4 \times 1 \times -10}}{2 \times 1}\) or \(\dfrac{9 \pm 11}{2}\) | M1 | oe ft their 3-term quadratic which cannot be \(x^2 - 3x - 10\) |
| \(-1\) and 10 | A1 | must have both solutions |
Additional guidance
| \(-1\) and 10 without working | M3A1 |
| In the quadratic formula \(9^2\) is equivalent to \((-9)^2\) but do not accept \(-9^2\) unless recovered | |
| \(x^2 - 3x - 10 = 6x\) | M1 |
| \(x^2 + 3x - 10 = 0\) | M0dep |
| \((x + 5)(x - 2) = 0\) | M1 |
| If first M1 is awarded for 4 terms that are incorrectly simplified to 3 terms, the 2nd M1 can be awarded using the incorrect simplification | |
| eg \(x^2 + 2x - 5x - 10 = x^2 - 7x - 10\) | M1 |
| \(x^2 - 7x - 10 - 6x\ (= 0)\) | M1dep |