FM Nov 2024 p12 q03
4145
It is given that
\(\alpha + \beta + \gamma + \delta = 2,\)
\(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 3,\)
\(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = 4.\)
(a) Find the value of \(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta.\)
(b) Find the value of \(\alpha^2 \beta + \alpha^2 \gamma + \alpha^2 \delta + \beta^2 \alpha + \beta^2 \gamma + \beta^2 \delta + \gamma^2 \alpha + \gamma^2 \beta + \gamma^2 \delta + \delta^2 \alpha + \delta^2 \beta + \delta^2 \gamma.\)
(c) It is given that \(\alpha, \beta, \gamma, \delta\) are the roots of the equation
\(6x^4 - 12x^3 + 3x^2 + 2x + 6 = 0.\)
(i) Find the value of \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4.\)
(ii) Find the value of \(\alpha^5 + \beta^5 + \gamma^5 + \delta^5.\)
Solution
(a) We use the identity \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).
Substituting the given values, \(3 = 2^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).
Solving, \(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \frac{1}{2}\).
(b) We use the identity \(\alpha^2(\beta + \gamma + \delta) + \beta^2(\alpha + \gamma + \delta) + \gamma^2(\alpha + \beta + \delta) + \delta^2(\alpha + \beta + \gamma)\).
This simplifies to \(\alpha^2(2-\alpha) + \beta^2(2-\beta) + \gamma^2(2-\gamma) + \delta^2(2-\delta)\).
Substituting the given values, \(2(3) - 4 = 2\).
(c)(i) Using the polynomial equation, \(6S_4 - 12S_3 + 3S_2 + 2S_1 + 24 = 0\).
Substitute \(S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_4 = \frac{11}{6}\).
(c)(ii) Using the polynomial equation, \(6S_5 - 12S_4 + 3S_3 + 2S_2 + S_1 = 0\).
Substitute \(S_4 = \frac{11}{6}, S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_5 = -\frac{4}{3}\).
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