CAIE P1 2007 June — Question 4 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2007
SessionJune
Marks4
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TopicSolving quadratics and applications
TypeQuadratic in negative or reciprocal fractional powers
DifficultyStandard +0.3 This is a straightforward substitution problem where students let u = 1/x² to obtain 18u² + u - 4 = 0, then solve the quadratic and back-substitute. It requires only routine algebraic manipulation and is slightly easier than average since the substitution is fairly obvious and the quadratic factors or solves cleanly.
Spec1.02f Solve quadratic equations: including in a function of unknown
4 Find the real roots of the equation \(\frac { 18 } { x ^ { 4 } } + \frac { 1 } { x ^ { 2 } } = 4\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\times (x^4) \rightarrow 4x^4 - x^2 - 18 = 0\)M1 Recognition of quadratic in \(x^2\) or \(1 \div (x^2)\)
\((4x^2 - 9)(x^2 + 2) = 0\)DM1 Solution of quadratic
\(x = 1.5\) or \(x = -1.5\)A1, A1\(\sqrt{}\) [4] Positive root. For recognition of \((-\text{ve})\); the A1\(\sqrt{}\) assumes no other real answers 
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\times (x^4) \rightarrow 4x^4 - x^2 - 18 = 0$ | M1 | Recognition of quadratic in $x^2$ or $1 \div (x^2)$ |
| $(4x^2 - 9)(x^2 + 2) = 0$ | DM1 | Solution of quadratic |
| $x = 1.5$ or $x = -1.5$ | A1, A1$\sqrt{}$ [4] | Positive root. For recognition of $(-\text{ve})$; the A1$\sqrt{}$ assumes no other real answers |

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4 Find the real roots of the equation $\frac { 18 } { x ^ { 4 } } + \frac { 1 } { x ^ { 2 } } = 4$.

\hfill \mbox{\textit{CAIE P1 2007 Q4 [4]}}
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