(a) $b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k+2)$ | M1A1 | For attempting to use $b^2 - 4ac$ with the values of $a, b$ and $c$ from the given equation. Fully correct (unsimplified) expression for $b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k+2)$. The bracketing must be correct. You can accept with or without any inequality signs. Accept $a = 2, b = 2k, c = k+2 \Rightarrow b^2 - 4ac = (2k)^2 - 4 \times 2 \times (k+2)$

$b^2 - 4ac > 0 \Rightarrow 4k^2 - 4 \times 2 \times (k+2) > 0 \Rightarrow k^2 - 2k - 4 > 0$ | A1* | Full proof, no errors, this is a given answer. It must be stated or implied that $b^2 - 4ac > 0$. Do not accept recovery from poor or incorrect bracketing or incorrect inequalities. Do not accept the answer written down without seeing an intermediate line such as $4k^2 - 4 \times 2 \times (k+2) > 0 \Rightarrow k^2 - 2k - 4 > 0$. Or $4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0$. The inequality must have been seen at least once before the final line for this mark to have been awarded. Eg accept $D = 4k^2 - 8k - 8 \Rightarrow 4k^2 - 8k - 8 > 0 \Rightarrow k^2 - 2k - 4 > 0$

| | **(3 marks)**

(b) $k^2 - 2k - 4 = 0 \Rightarrow (k-1)^2 = 5$ | M1 | Attempt to solve the given 3 term quadratic (=0) by formula or completing the square. Do NOT accept an attempt to factorise in this question. If the formula is given it must be correct. It can be implied by seeing either $\frac{-(−2) \pm \sqrt{(−2)^2 − 4 \times 1 \times −4}}{2 \times 1}$ or $− − 2 \pm \sqrt{−2^2 − 4 \times 1 \times −4} / 2 \times 1$. If completing the square is used it can be implied by $(k-1)^2 \pm 1-4 = 0 \Rightarrow k = ...$

$k = 1 \pm \sqrt{5}$ oe | A1 | Obtains critical values of $1 \pm \sqrt{5}$. Accept $\frac{2 \pm \sqrt{20}}{2}$

$k > 1 + \sqrt{5}, \quad k < 1 - \sqrt{5}$ | dM1A1 | Outside of their values chosen. It is dependent upon the previous M mark having been awarded. States $k >$ their largest value. $k <$ their smallest value. Do not award simply for a diagram or a table- they must have chosen their "outside regions". Correct answer only. Accept $k > 1 + \sqrt{5}$ or $k < 1 - \sqrt{5}, k > 1 + \sqrt{5} k < 1 - \sqrt{5}$, $(-\infty, 1-\sqrt{5}) \cup (1+\sqrt{5}, \infty)$ but not $k > 1 + \sqrt{5}$ and $k < 1 - \sqrt{5}, 1 + \sqrt{5} < k < 1 - \sqrt{5}$

| | **(4 marks)**
| | **(7 marks)**

**Alt (a)** $b^2 > 4ac \Rightarrow (2k)^2 > 4 \times 2 \times (k+2)$ | M1A1 | 

$\Rightarrow k^2 - 2k - 4 > 0$ | A1* |

| | **(3 marks)**

**Notes for Question 8 continued**

Also accept exact alternatives as a simplified form is not explicitly asked for in the question. Accept versions such as $k > \frac{2 + \sqrt{20}}{2}$ or $k < \frac{2 - \sqrt{20}}{2}$

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