| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| 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), leading to a simple quadratic equation in p, followed by routine substitution to solve. It requires only standard recall and basic algebraic manipulation with no problem-solving insight needed. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| \(b^2-4ac=4p^2-4(3p+4)=4p^2-12p-16\,(=0)\) or \((x+p)^2-p^2+(3p+4)=0\Rightarrow p^2-3p-4\,(=0)\) | M1, A1 | For use of \(b^2-4ac\) or a full attempt to complete the square leading to a 3TQ in \(p\). May use \(b^2=4ac\). One of \(b\) or \(c\) must be correct. For a correct 3TQ in \(p\). Condone missing "\(=0\)" but all 3 terms must be on one side |
| \((p-4)(p+1)=0\) | M1 | For attempt to solve their 3TQ leading to \(p=\ldots\) |
| \(p=(-1\text{ or })\,4\) | A1c.s.o. | For \(p=4\) (ignore \(p=-1\)). \(b^2=4ac\) leading to \(p^2=4(3p+4)\) and then "spotting" \(p=4\) scores 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=\frac{-b}{2a}\) or \((x+p)(x+p)=0\Rightarrow x=\ldots\) | M1 | For a full method leading to a repeated root \(x=\ldots\) |
| \(x\,(=-p)=-4\) | A1f.t. | For \(x=-4\) (- their \(p\)) |
## Question 8:
### Part (a):
| $b^2-4ac=4p^2-4(3p+4)=4p^2-12p-16\,(=0)$ or $(x+p)^2-p^2+(3p+4)=0\Rightarrow p^2-3p-4\,(=0)$ | M1, A1 | For use of $b^2-4ac$ or a full attempt to complete the square leading to a 3TQ in $p$. May use $b^2=4ac$. One of $b$ or $c$ must be correct. For a correct 3TQ in $p$. Condone missing "$=0$" but all 3 terms must be on one side |
| $(p-4)(p+1)=0$ | M1 | For attempt to solve their 3TQ leading to $p=\ldots$ |
| $p=(-1\text{ or })\,4$ | A1c.s.o. | For $p=4$ (ignore $p=-1$). $b^2=4ac$ leading to $p^2=4(3p+4)$ and then "spotting" $p=4$ scores 4/4 |
### Part (b):
| $x=\frac{-b}{2a}$ or $(x+p)(x+p)=0\Rightarrow x=\ldots$ | M1 | For a full method leading to a repeated root $x=\ldots$ |
| $x\,(=-p)=-4$ | A1f.t. | For $x=-4$ (- their $p$) |
---8. The equation $x ^ { 2 } + 2 p x + ( 3 p + 4 ) = 0$, where $p$ is a positive constant, has equal roots.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$.
\item For this value of $p$, solve the equation $x ^ { 2 } + 2 p x + ( 3 p + 4 ) = 0$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2006 Q8 [6]}}