June 2024 Paper 1 Q14
14
(a) Write \(\quad 2x^2 - 16x + 7 \quad\) in the form \(\quad k(x + m)^2 + n \quad\) where \(k\), \(m\) and \(n\) are integers. [3 marks]
(b) Solve \(\quad (x - 1)^2 - 5 = 0\) [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(2(x^2 - 8x + \ldots\ldots\) \(2(x - 4)^2 + \ldots\ldots\) | M1 | oe |
| \(2[(x - 4)^2 - 4^2] + \ldots\ldots\) or \(2[(x - 4)^2 - 16] + \ldots\ldots\) or \(2\left[(x - 4)^2 - 16 + \dfrac{7}{2}\right]\) or \(2\left[(x - 4)^2 - \dfrac{25}{2}\right]\) or \(2(x - 4)^2 - 2 \times \dfrac{25}{2}\) or \(2(x - 4)^2 - 32 + 7\) | M1dep | oe the bracket is after the \(4^2\) and the 16 here. If they put something else inside the bracket it is incorrect unless it is equivalent to one of the fully complete versions listed |
| \(2(x - 4)^2 - 25\) | A1 | |
| Alternative method 2 – using identities | ||
| \(kx^2 + 2kmx + km^2 + n\) | M1 | RHS oe |
| \(k = 2\) and \(2km = -16\) and \(km^2 + n = 7\) | M1dep | oe |
| \(2(x - 4)^2 - 25\) | A1 | |
Additional guidance
| Condone answer followed by \(k = 2 \quad m = -4 \quad n = -25\) | M2A1 |
| \(k = 2 \quad m = -4 \quad n = -25\) without correct answer is penalised | M2A0 |
| Answer | Mark | Comments |
|---|---|---|
| \((x =)\ 1 \pm \sqrt{5}\) | B1 |
Additional guidance
| Condone roots written separately (and/or are both fine) | |
| Do not condone \(1 + \sqrt{5}\) on its own or \(1 - \sqrt{5}\) on its own | |
| Do not accept oe answers – must be simplified |