Edexcel C1 2013 June — Question 10 8 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2013
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimultaneous equations
TypeLine intersecting quadratic curve
DifficultyModerate -0.8 This is a routine C1 question testing standard techniques: substitution to eliminate a variable, then using the discriminant condition for equal roots (b²=4ac), followed by straightforward solving. The algebra is mechanical with no conceptual challenges or problem-solving insight required—easier than average A-level questions.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions
  1. Given the simultaneous equations
$$\begin{aligned} & 2 x + y = 1 \\ & x ^ { 2 } - 4 k y + 5 k = 0 \end{aligned}$$ where \(k\) is a non zero constant,
  1. show that $$x ^ { 2 } + 8 k x + k = 0$$ Given that \(x ^ { 2 } + 8 k x + k = 0\) has equal roots,
  2. find the value of \(k\).
  3. For this value of \(k\), find the solution of the simultaneous equations.
Question 10:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^2 - 4k(1-2x) + 5k(= 0)\)M1 Makes \(y\) the subject from first equation and substitutes into second equation, or eliminates \(y\) by correct method
So \(x^2 + 8kx + k = 0\) *A1cso Correct completion to printed answer. No incorrect statements.
(2 marks)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((8k)^2 - 4k\)M1 A1 M1: Use of \(b^2-4ac\) (could be in quadratic formula or inequality, \(=0\) not needed yet). Must be correct substitution with no \(x\)'s. \(8k^2-4k=0\) is M0. A1: Correct expression — do not condone missing brackets unless implied by later work e.g. \((8k)^2 > 4k\)
\(k = \frac{1}{16}\) (oe)A1 cso — ignore any reference to \(k=0\) but no contradictory earlier statements. Fully correct solution with no errors.
Way 2 (Equal roots):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\Rightarrow x^2 + 8kx + k = (x+\sqrt{k})^2\)M1 A1 M1: Correct strategy for equal roots. A1: Correct equation
\(\Rightarrow 8k = 2\sqrt{k}\)
\(k = \frac{1}{16}\) (oe)A1 cso (ignore reference to \(k=0\))
Way 3 (Complete the Square):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^2+8kx+k = (x+4k)^2 - 16k^2 + k \Rightarrow 16k^2 - k = 0\)M1 A1 M1: \((x \pm 4k)^2 \pm p \pm k\), \(p \neq 0\). A1: Correct equation
\(k = \frac{1}{16}\) (oe)A1 cso (ignore reference to \(k=0\))
(3 marks)
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^2 + \frac{1}{2}x + \frac{1}{16} = 0\) so \((x+\frac{1}{4})^2 = 0 \Rightarrow x = ...\)M1 Substitutes their \(k\) into given quadratic and attempts to solve 2 or 3 term quadratic as far as \(x=\) (may be implied by substitution into quadratic formula) OR substitutes \(k\) into second equation and solves simultaneously for \(x\)
\(x = -\frac{1}{4}\), \(y = 1\frac{1}{2}\)A1 A1 First A1: one answer correct. Second A1: both answers correct
Special case: \(x^2 + \frac{1}{2}x + \frac{1}{16} = 0 \Rightarrow x = -\frac{1}{4}, \frac{1}{4} \Rightarrow y = 1\frac{1}{2}, \frac{1}{2}\) allow M1A1A0. (3 marks) [8 marks total]
## Question 10:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - 4k(1-2x) + 5k(= 0)$ | M1 | Makes $y$ the subject from first equation and substitutes into second equation, or eliminates $y$ by correct method |
| So $x^2 + 8kx + k = 0$ * | A1cso | Correct completion to printed answer. No incorrect statements. |

**(2 marks)**

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(8k)^2 - 4k$ | M1 A1 | M1: Use of $b^2-4ac$ (could be in quadratic formula or inequality, $=0$ not needed yet). Must be correct substitution with no $x$'s. $8k^2-4k=0$ is M0. A1: Correct expression — do not condone missing brackets unless implied by later work e.g. $(8k)^2 > 4k$ |
| $k = \frac{1}{16}$ (oe) | A1 | cso — ignore any reference to $k=0$ but no contradictory earlier statements. Fully correct solution with no errors. |

**Way 2 (Equal roots):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\Rightarrow x^2 + 8kx + k = (x+\sqrt{k})^2$ | M1 A1 | M1: Correct strategy for equal roots. A1: Correct equation |
| $\Rightarrow 8k = 2\sqrt{k}$ | | |
| $k = \frac{1}{16}$ (oe) | A1 | cso (ignore reference to $k=0$) |

**Way 3 (Complete the Square):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2+8kx+k = (x+4k)^2 - 16k^2 + k \Rightarrow 16k^2 - k = 0$ | M1 A1 | M1: $(x \pm 4k)^2 \pm p \pm k$, $p \neq 0$. A1: Correct equation |
| $k = \frac{1}{16}$ (oe) | A1 | cso (ignore reference to $k=0$) |

**(3 marks)**

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + \frac{1}{2}x + \frac{1}{16} = 0$ so $(x+\frac{1}{4})^2 = 0 \Rightarrow x = ...$ | M1 | Substitutes their $k$ into given quadratic and attempts to solve 2 or 3 term quadratic as far as $x=$ (may be implied by substitution into quadratic formula) OR substitutes $k$ into second equation and solves simultaneously for $x$ |
| $x = -\frac{1}{4}$, $y = 1\frac{1}{2}$ | A1 A1 | First A1: one answer correct. Second A1: both answers correct |

**Special case:** $x^2 + \frac{1}{2}x + \frac{1}{16} = 0 \Rightarrow x = -\frac{1}{4}, \frac{1}{4} \Rightarrow y = 1\frac{1}{2}, \frac{1}{2}$ allow M1A1A0. **(3 marks) [8 marks total]**

---
\begin{enumerate}
  \item Given the simultaneous equations
\end{enumerate}

$$\begin{aligned}
& 2 x + y = 1 \\
& x ^ { 2 } - 4 k y + 5 k = 0
\end{aligned}$$

where $k$ is a non zero constant,\\
(a) show that

$$x ^ { 2 } + 8 k x + k = 0$$

Given that $x ^ { 2 } + 8 k x + k = 0$ has equal roots,\\
(b) find the value of $k$.\\
(c) For this value of $k$, find the solution of the simultaneous equations.\\

\hfill \mbox{\textit{Edexcel C1 2013 Q10 [8]}}
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