Question

Challenger question for maths genius!
It is easy too but for me hard only a little ,need oF understanding

Please solve (iii) part

Answer 1

$Answer \: \\ \\ Given \: \\ \alpha \: \: \: and \: \: \: \beta \: \: are \: the \: zeros \: of \: \: x {}^{2} - x - 2 \: \: \\ \\ \alpha + \beta = \frac{ - ( -1 )}{1} \: \: \: and \: \: \: \alpha \beta = \frac{ - 2}{1} \\ \\ \alpha + \beta = 1 \: \: \: and \: \: \: \alpha \beta = - 2 \\ \\ let \: the \: required \: quadratic \: polynomial \: be \: \: p(x) \\ \\ p(x) = x {}^{2} - ( \frac{ \alpha {}^{2} }{ \beta {}^{2} } + \frac{ \beta {}^{2} }{ \alpha {}^{2} } )x + ( \frac{ \alpha {}^{2} }{ \beta {}^{2} } \times \frac{ \beta {}^{2} }{ \alpha {}^{2} } ) \\ \\ p(x) = x {}^{2} - ( \frac{ \alpha {}^{2} }{ \beta {}^{2} } + \frac{ \beta {}^{2} }{ \alpha {}^{2} } ) x+ 1 \\ \\ p(x) = x {}^{2} - ( \frac{ \alpha {}^{4} + \beta {}^{4} }{ \alpha {}^{2} \beta {}^{2} } )x + 1 \\ \\ Now \: \: \: \\ \\ \alpha {}^{4} + \beta {}^{4} = (( \alpha + \beta ) {}^{2} - 2 \alpha \beta ) {}^{2} - 2( \alpha \beta ) {}^{2} \\ \\ \alpha {}^{4} + \beta {}^{4} = ((1) {}^{2} - 2( - 2)) {}^{2} - 2( - 2) {}^{2} \\ \\ \alpha {}^{4} + \beta {}^{4} = (1 + 4) {}^{2} - 8 \\ \\ \alpha {}^{4} + \beta {}^{4} = 25 - 8 \\ \\ \alpha {}^{4} + \beta {}^{4} = 17 \\ \\ and \: \: \: \: \: \: \: \: \: \: ( \alpha \beta ) {}^{2} = ( - 2) {}^{2} = 4 \\ \\ so \: the \: required \: quadratic \: polynomial \: is \: \\ \\ p(x) = x {}^{2} - ( \frac{17}{4} )x + 1 \\ \\ p(x) = (4x {}^{2} - 17x + 4) \frac{1}{4} \\ \\ and \: the \: required \: quadratic \: equation \: is \\ \\ 4x {}^{2} - 17x + 4 = 0$