## Question 8:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2x-3)(x+1)=0$ | M1 | Correct method to find roots – see appendix 1 |
| $x=\frac{3}{2}, x=-1$ | A1 | Correct roots |
| Graph: correct positive quadratic, minimum in correct quadrant, $x$-intercepts correctly labelled | A1ft | Good curve: correct shape, symmetrical positive quadratic; minimum point in the correct quadrant for their roots **(ft)**; their $x$ intercepts correctly labelled **(ft)** |
| $y$ intercept at $(0,-3)$ | B1 | Must have a graph |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x<-1, x>\frac{3}{2}$ | M1 | Chooses the "outside region". **If restarted**, fully correct method for solving a quadratic inequality including choosing "outside region" needed for M1 |
| | A1ft | Follow through $x$-values in (i). Allow "$x<-1, x>\frac{3}{2}$", "$x<-1$ or $x>\frac{3}{2}$" but do not allow "$x<-1$ **and** $x>\frac{3}{2}$". **NB** e.g. $-1>x>\frac{3}{2}$ scores M1A0. Must be strict inequalities for A mark |

### Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $b^2-4ac=1^2-4\times2\times-(3+k)$ | M1 | Rearrangement and use of $b^2-4ac<0$, must involve 3 and $k$ in constant term (not $3k$). **Alt:** M1 Attempt to find turning point and form inequality $k < y_{min}$ |
| $25+8k<0$ | A1 | $p+8k<0$ oe found, any constant $p$. $p$ need not be simplified. **A1** turning point correct $\left(\frac{1}{4},-\frac{25}{8}\right)$ |
| $k<-\frac{25}{8}$ | A1 | Correct final answer. **If M0** (either scheme) SC B1 $k=-\frac{25}{8}$ or $k>-\frac{25}{8}$ seen |

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