Edexcel C3 — Question 4 9 marks

Exam BoardEdexcel
ModuleC3 (Core Mathematics 3)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypeSimplify then solve or apply
DifficultyStandard +0.3 This is a standard C3 algebraic fractions question requiring common denominator manipulation and completing the square. Part (a) involves routine algebraic manipulation (5 marks suggests multiple steps but no novel insight). Part (b) requires recognizing that a quadratic with negative discriminant has no real roots—a standard technique. Slightly above average difficulty due to the algebraic complexity, but follows predictable C3 patterns without requiring creative problem-solving.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02y Partial fractions: decompose rational functions

  1. Express $$\frac{x-10}{(x-3)(x+4)} - \frac{x-8}{(x-3)(2x-1)}$$ as a single fraction in its simplest form. [5]
  2. Hence, show that the equation $$\frac{x-10}{(x-3)(x+4)} - \frac{x-8}{(x-3)(2x-1)} = 1$$ has no real roots. [4]

AnswerMarks
(a) \(= \frac{(x-10)(2x-1) - (x-8)(x+4)}{(x-3)(x+4)(2x-1)}\)M1 A1
\(= \frac{x^2-17x+42}{(x-3)(x+4)(2x-1)}\)A1
\(= \frac{(x-14)(x-3)}{(x-3)(x+4)(2x-1)} = \frac{x-14}{(x+4)(2x-1)}\)M1 A1
(b) \(\frac{x-14}{(x+4)(2x-1)} = 1\), \(x - 14 = 2x^2 + 7x - 4\)M1
\(x^2 + 3x + 5 = 0\)A1
\(b^2 - 4ac = 9 - 20 = -11\)
AnswerMarks Guidance
\(b^2 - 4ac < 0 \therefore\) no real rootsM1 A1 (9 marks)
**(a)** $= \frac{(x-10)(2x-1) - (x-8)(x+4)}{(x-3)(x+4)(2x-1)}$ | M1 A1 |
$= \frac{x^2-17x+42}{(x-3)(x+4)(2x-1)}$ | A1 |
$= \frac{(x-14)(x-3)}{(x-3)(x+4)(2x-1)} = \frac{x-14}{(x+4)(2x-1)}$ | M1 A1 |

**(b)** $\frac{x-14}{(x+4)(2x-1)} = 1$, $x - 14 = 2x^2 + 7x - 4$ | M1 |
$x^2 + 3x + 5 = 0$ | A1 |
$b^2 - 4ac = 9 - 20 = -11$
$b^2 - 4ac < 0 \therefore$ no real roots | M1 A1 | (9 marks)
\begin{enumerate}[label=(\alph*)]
\item Express
$$\frac{x-10}{(x-3)(x+4)} - \frac{x-8}{(x-3)(2x-1)}$$
as a single fraction in its simplest form. [5]

\item Hence, show that the equation
$$\frac{x-10}{(x-3)(x+4)} - \frac{x-8}{(x-3)(2x-1)} = 1$$
has no real roots. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C3  Q4 [9]}}