Question

pls answer quickly.......​

Answer 1

$HEYA \: \\ \\ required \: quadratic \: equation \: is \: \\ \\ x {}^{2} + (sum \: of \: roots \: )x + (product \: of \: roots) = 0 \\ \\ here \: \: sum \: of \: roots \: = ( \alpha {}^{2} + \beta {}^{2} ) \div \alpha \beta \\ and \\ product \: of \: roots \: = \: \: 1 \\ \\ ( \alpha + \beta ) = - p \: \: \: (given \: ) \\ cubing \: on \: both \: sides \: we \: have \\ \\ ( \alpha + \beta ) {}^{3} = - p {}^{3} \\ \\ \alpha {}^{3} + \beta {}^{3} + 3 \alpha \beta ( \alpha + \beta ) = - p {}^{3} \\ \\ 3 \alpha \beta = (- p {}^{3} - q) \div ( - p) \\ \\ 3 \alpha \beta = (p {}^{3} + q) \div p \\ \\ \alpha \beta = (p {}^{3} + q) \div 3p .....(i)\\ \\ (\alpha + \beta ) {}^{2} = p {}^{2} \\ \\ \alpha {}^{2} + \beta {}^{2} = (p {}^{2} - 2 \alpha \beta ) \\ \\ divide \: \: both \: by \: \alpha \beta \\ \\ ( \alpha {}^{2} + \beta {}^{2} ) \div ( \alpha \beta ) = (p {}^{2} \div ( \alpha \beta) - 2) \\ \\ ( \alpha {}^{2} + \beta {}^{2} ) \div ( \alpha \beta ) = (p { }^{3} - 2 q) \div (p {}^{3} + q) \\ \\ so \: required \: quadratic \: equation \: is \\ \\ x {}^{2}(p {}^{3} + q) + (p {}^{3} - 2 q)x + (p {}^{3} + q) = 0$