SPS SPS FM Pure 2023 October — Question 4 10 marks

Exam BoardSPS
ModuleSPS FM Pure (SPS FM Pure)
Year2023
SessionOctober
Marks10
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyStandard +0.8 This is a Further Maths question requiring a non-standard substitution to transform roots, followed by applying Vieta's formulas to the new equation. The substitution x = 1/√u is not routine and requires careful algebraic manipulation. Part (ii) demands recognizing that the transformed roots relate to 1/α², 1/β², 1/γ² and applying symmetric function formulas. While systematic, it requires more insight than standard root transformation questions and involves multiple sophisticated steps.
Spec4.05b Transform equations: substitution for new roots

4. The cubic equation \(x ^ { 3 } + 3 x ^ { 2 } + 2 = 0\) has roots \(\alpha , \beta\) and \(\gamma\).
  1. Use the substitution \(x = \frac { 1 } { \sqrt { u } }\) to show that \(4 u ^ { 3 } + 12 u ^ { 2 } + 9 u - 1 = 0\).
  2. Hence find the values of \(\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } }\) and \(\frac { 1 } { \alpha ^ { 2 } \beta ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } \alpha ^ { 2 } }\).
    [0pt]

4. The cubic equation $x ^ { 3 } + 3 x ^ { 2 } + 2 = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
(i) Use the substitution $x = \frac { 1 } { \sqrt { u } }$ to show that $4 u ^ { 3 } + 12 u ^ { 2 } + 9 u - 1 = 0$.\\
(ii) Hence find the values of $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } }$ and $\frac { 1 } { \alpha ^ { 2 } \beta ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } \alpha ^ { 2 } }$.\\[0pt]
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\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q4 [10]}}