Edexcel C1 Specimen — Question 8 11 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCompleting the square and sketching
TypeQuadratic with equal roots
DifficultyEasy -1.2 This is a straightforward C1 completing the square question with standard parts: finding constants (routine algebra), showing no real roots (direct observation from completed square form), finding k for equal roots (discriminant = 0 or completed square = 0), and sketching a parabola. All parts are textbook exercises requiring only direct application of learned techniques with no problem-solving insight needed.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials

Given that $$x^2 + 10x + 36 = (x + a)^2 + b$$ where \(a\) and \(b\) are constants,
  1. find the value of \(a\) and the value of \(b\). [3]
  2. Hence show that the equation \(x^2 + 10x + 36 = 0\) has no real roots. [2]
The equation \(x^2 + 10x + k = 0\) has equal roots.
  1. Find the value of \(k\). [2]
  2. For this value of \(k\), sketch the graph of \(y = x^2 + 10x + k\), showing the coordinates of any points at which the graph meets the coordinate axes. [4]

Part (a)
AnswerMarks Guidance
\(a = 5\) and \((x+5)^2 - 25 + 36\) and \(b = 11\)B1, M1 A1 (3 marks)
Part (b)
AnswerMarks Guidance
\(b^2 - 4ac = 100 - 144 < 0\), therefore no real rootsM1 A1 (2 marks)
Part (c)
AnswerMarks Guidance
Equal roots if \(b^2 - 4ac = 0\) and \(4k = 100\) and \(k = 25\)M1 A1 (2 marks)
Part (d)
AnswerMarks Guidance
Shape, positionB1 B1
\((-5, 0)\) \((0, 25)\)B1 B1ft (4 marks)
Total: 11 marks
## Part (a)
$a = 5$ and $(x+5)^2 - 25 + 36$ and $b = 11$ | B1, M1 A1 | (3 marks)

## Part (b)
$b^2 - 4ac = 100 - 144 < 0$, therefore no real roots | M1 A1 | (2 marks)

## Part (c)
Equal roots if $b^2 - 4ac = 0$ and $4k = 100$ and $k = 25$ | M1 A1 | (2 marks)

## Part (d)
Shape, position | B1 B1 |
$(-5, 0)$ $(0, 25)$ | B1 B1ft | (4 marks)

**Total: 11 marks**

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Given that
$$x^2 + 10x + 36 = (x + a)^2 + b$$
where $a$ and $b$ are constants,

\begin{enumerate}[label=(\alph*)]
\item find the value of $a$ and the value of $b$. [3]
\item Hence show that the equation $x^2 + 10x + 36 = 0$ has no real roots. [2]
\end{enumerate}

The equation $x^2 + 10x + k = 0$ has equal roots.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $k$. [2]
\item For this value of $k$, sketch the graph of $y = x^2 + 10x + k$, showing the coordinates of any points at which the graph meets the coordinate axes. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q8 [11]}}