Higher June 2024 Paper 3 Q24
24 Write \(\quad 2x^2 - 12x + 7 \quad\) in the form \(\quad d(x + e)^2 + f\)
where \(d\), \(e\) and \(f\) are integers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(dx^2\) or \(2dex\) or \(de^2\) | M1 | |
| \(dx^2 + 2dex + de^2 + f\) | M1dep | |
| \(2(x - 3)^2 - 11\) or \(d = 2,\ e = -3,\ f = -11\) | A1 | SC2 \(2(x - 6)^2 - 29\) SC1 \(2(x - 6)^2 + k \quad k \ne -29\) SC1 \(2(x + 6)^2 - 29\) SC1 \(2(x + 3)^2 + k\) SC1 \((x - 3)^2 - 2\) |
| Alternative method 2 | ||
| \(2(x^2 \ldots)\) or \(d = 2\) | M1 | |
| \(2\left(x^2 - 6x + \dfrac{7}{2}\right)\) or \(2(x^2 - 6x) + 7\) or \(2(x - 3)^2 + k\) | M1dep | \(k \ne -11\) |
| \(2(x - 3)^2 - 11\) or \(d = 2,\ e = -3,\ f = -11\) | A1 | SC2 \(2(x - 6)^2 - 29\) SC1 \(2(x - 6)^2 + k \quad k \ne -29\) SC1 \(2(x + 6)^2 - 29\) SC1 \(2(x + 3)^2 + k\) SC1 \((x - 3)^2 - 2\) |