Question

if alpha and beta are the zeroes of polynomial 3xsquare-5x-2 then find the value of alpha square/beta +beta square/alpha

Answer 1

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Answer 2

$HEYA \: \\ \\ GIVEN \: \: QUESTION \: \: Is \: \: \\ \\ f(x) = 3x {}^{2} - 5x - 2 \\ \alpha \: \: \: and \: \: \: \beta \: \: are \: the \: zeros \: of \: \: f(x) \\ \\ \alpha + \beta = \frac{5}{3} \: \: \: and \: \: \: \alpha \beta = \frac{ - 2}{3} \\ \\ \frac{ \alpha {}^{2} }{ \beta } + \frac{ \beta {}^{2} }{ \alpha } = \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } = \frac{( \alpha + \beta ) {}^{3} - 3 \alpha \beta ( \alpha + \beta )}{ \alpha \beta } \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } = \frac{( \frac{5}{3} ) {}^{3} - 3 (\frac{ - 2}{3} )( \frac{5}{3} )}{ (\frac{ - 2}{3}) } \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } = \frac{ \frac{125}{27} + \frac{10}{3} }{ \frac{ - 2}{3} } \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} } { \alpha \beta } = \frac{ \frac{125}{27} + \frac{90}{27} }{ \frac{ - 2}{3} } \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } = \frac{ \frac{215}{27} }{ \frac{ - 2}{3} } \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } = \frac{215 \times 3}{27 \times - 2} \\ \\ \\ \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha \beta } = \frac{215}{ - 18}$