OCR C1 2005 June — Question 4 5 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Year2005
SessionJune
Marks5
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TopicSolving quadratics and applications
TypeQuadratic in higher integer powers
DifficultyStandard +0.3 This is a straightforward substitution question where students let y = x³ to get y² + 26y - 27 = 0, solve the quadratic, then find x. While it requires recognizing the substitution pattern and solving a quadratic, it's a standard C1 technique with no conceptual challenges—slightly easier than average since the substitution is obvious and the arithmetic is clean.
Spec1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem
4 Solve the equation \(x ^ { 6 } + 26 x ^ { 3 } - 27 = 0\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(k = x^3\)*M1 Attempt substitution to obtain a quadratic
\(k^2 + 26k - 27 = 0\)A1
\(k = -27, 1\)A1
DM1Attempt cube root
\(x = -3, 1\)A1 5 No extras. SR: \(x=1\) seen www B1; \(x=-3\) seen www B1 
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $k = x^3$ | *M1 | Attempt substitution to obtain a quadratic |
| $k^2 + 26k - 27 = 0$ | A1 | |
| $k = -27, 1$ | A1 | |
| | DM1 | Attempt cube root |
| $x = -3, 1$ | A1 **5** | No extras. SR: $x=1$ seen www **B1**; $x=-3$ seen www **B1** |

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4 Solve the equation $x ^ { 6 } + 26 x ^ { 3 } - 27 = 0$.

\hfill \mbox{\textit{OCR C1 2005 Q4 [5]}}
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