Higher November 2018 Paper 1 Q12
12 Solve \(\quad x^2 - x - 12 = 0\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x + a)(x + b)\) | M1 | where \(ab = \pm 12\) or \(a + b = -1\) |
| \((x - 4)(x + 3)\) | A1 | |
| 4 and \(-3\) | A1 | SC1 4 or \(-3\) with no or one incorrect answer |
| Alternative method 2 | ||
| \(\dfrac{(--)1 \pm \sqrt{((-)1)^2 - 4(1)(-12)}}{2(1)}\) or \(\dfrac{1 \pm \sqrt{1 + 48}}{2}\) or \(\dfrac{1 \pm \sqrt{49}}{2}\) | M1 | oe allow one sign error |
| \(\dfrac{(--)1 \pm \sqrt{((-)1)^2 - 4(1)(-12)}}{2(1)}\) or \(\dfrac{1 \pm \sqrt{1 + 48}}{2}\) or \(\dfrac{1 \pm \sqrt{49}}{2}\) | A1 | oe fully correct |
| 4 and \(-3\) | A1 | SC1 4 or \(-3\) with no or one incorrect answer |
| Alternative method 3 | ||
| \(\left(x - \dfrac{1}{2}\right)^2 \ldots\) | M1 | |
| \(\left(x - \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2 - 12\ (= 0)\) | A1 | oe equation |
| 4 and \(-3\) | A1 | SC1 4 or \(-3\) with no or one incorrect answer |
Additional guidance
| 4 and \(-3\) with no working | M1A1A1 |
| M1 can be scored amongst incorrect attempts to factorise | |
| Condone trailing bracket missing eg \(\ (x - 4)(x + 3\) | M1A1 |