| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Find k for equal roots |
| Difficulty | Moderate -0.8 This is a straightforward application of the discriminant condition for equal roots (b² - 4ac = 0), followed by solving a simple quadratic by factorisation. Both parts are routine C1 procedures requiring only standard recall and algebraic manipulation, making it easier than average but not trivial. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \((m+4)^2 = m^2 + 8m + 16\) | B1 | Condone \(4m + 4m\) |
| \(b^2 - 4ac = (m+4)^2 - 4(4m+1) = 0\) | M1 | \(b^2 - 4ac\) (attempted and involving m's and no x's) or \(b^2 - 4ac = 0\) stated |
| \(m^2 + 8m + 16 - 16m - 4 = 0 \Rightarrow m^2 - 8m + 12 = 0\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((m-2)(m-6) = 0\) | M1 | Attempt at factors or quadratic formula |
| \(m = 2, m = 6\) | A1 | 2 |
## 4(a)
$(m+4)^2 = m^2 + 8m + 16$ | B1 | Condone $4m + 4m$
$b^2 - 4ac = (m+4)^2 - 4(4m+1) = 0$ | M1 | $b^2 - 4ac$ (attempted and involving m's and no x's) or $b^2 - 4ac = 0$ stated
$m^2 + 8m + 16 - 16m - 4 = 0 \Rightarrow m^2 - 8m + 12 = 0$ | A1 | 3 | AG (be convinced – all working correct- = 0 appearing more than right at the end)
## 4(b)
$(m-2)(m-6) = 0$ | M1 | Attempt at factors or quadratic formula
$m = 2, m = 6$ | A1 | 2 | SC B1 for 2 or 6 only without working
---4 The quadratic equation $x ^ { 2 } + ( m + 4 ) x + ( 4 m + 1 ) = 0$, where $m$ is a constant, has equal roots.
\begin{enumerate}[label=(\alph*)]
\item Show that $m ^ { 2 } - 8 m + 12 = 0$.
\item Hence find the possible values of $m$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2006 Q4 [5]}}