Question

answer this polynomial ​

Answer 1

$let \: p(x) be \: the \: given \: quadratic \: polynomial \\ then \: p(x) = a {x}^{2} + bx + c \\ since \: \alpha \: and \: \beta \: are \: the \: zeros \: of \: p(x) \\ therefore \: \alpha \: and \: \beta \: are \: the \: roots \: of \: quaratic \:equation \: a {x}^{2} + bx + c = 0 \\ which\: means \: \alpha + \beta = - \frac{ b}{a} \: (equation \: 1) \\ and \: \alpha \beta \: = \frac{c}{a} \: (equation \: 2) \\ now \: from \: question \\ ( { \alpha }^{ - 1} + \frac{1}{{ \alpha }^{ - 1}} )( { \beta }^{ - 1} + \frac{1}{{ \beta }^{ - 1}} ) = ( \frac{1}{ \alpha } + \alpha )( \frac{1}{ \beta } + \beta ) \\ = \frac{1 + { \alpha }^{2} }{ \alpha } \times \frac{1 + { \beta }^{2} }{ \beta } \\ = \frac{ { (\alpha }^{2} + 1)( { \beta }^{2} + 1)}{ \alpha \beta } \\ = \frac{ { \alpha }^{2} { \beta }^{2} + { \alpha }^{2} + { \beta }^{2} + 1}{ \alpha \beta } \\ = \frac{{( \alpha \beta )}^{2} + {( \alpha + \beta )}^{2} - 2 \alpha \beta + 1 }{ \alpha \beta } \\ = \frac{ \frac{ {c}^{2} }{ {a}^{2} } + ( - {\frac{b}{a} )}^{2} - \frac{2c}{a} + 1}{ \frac{c}{a} } \: (from \: equation \: 1 \: and \: 2) \\ = \frac{a}{ c} \times ( \frac{ {c}^{2} }{ {a}^{2} } + \frac{ {b}^{2} }{ {a}^{2} } - \frac{2c}{a} + 1) \\ = \frac{a}{c} \times ( \frac{ {b}^{2} + {c}^{2} }{ {a}^{2} } - \frac{2c}{a} + 1) \\ = \frac{a( {b}^{2} + {c}^{2} )}{ {a}^{2}c } - 2 + \frac{a}{c} \\ = \frac{ {b}^{2} + {c}^{2} }{ac} + \frac{ - 2c + a}{c} \\ = \frac{1}{c} ( \frac{ {b}^{2} + {c}^{2} }{a} + ( - 2c + a)) \\ = \frac{1}{c} ( \frac{ {b}^{2} + {c}^{2} + a( - 2c + a)}{a} ) \\ = \frac{1}{c} ( \frac{ {b}^{2} + {c}^{2} - 2ac + {a}^{2} }{a} ) \\ = \frac{ {a}^{2} + {c}^{2} - 2ca + {b}^{2} }{ac} \\ = \frac{ {(a - c)}^{2} + {b}^{2} }{ac}$