Question

plz solve guys question no.18 plz plz

Answer 1

${x}^{2} - 2x + 5 \\ \alpha + \beta = 2 \\ \alpha \beta = 5 \\ \frac{ \beta }{2 \alpha + 1 } + \frac{ \alpha }{2 \beta + 1} \\ = \frac{ \beta (2 \beta + 1) + \alpha (2 \alpha + 1)}{(2 \alpha + 1)(2 \beta + 1)} \\ = \frac{2 { \beta }^{2} + \beta + 2 { \alpha }^{2} + \alpha }{4 { \alpha } \beta + 2 \alpha + 2 \beta + 1} \\ = \frac{( \alpha + \beta) + 2( { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta ) - 4 \alpha \beta }{4 \alpha \beta + 2( \alpha + \beta ) + 1} \\ = \frac{( \alpha + \beta) + 2( { \alpha + \beta )}^{2} - 4 \alpha \beta }{4 \alpha \beta + 2( \alpha + \beta ) + 1} \\ = \frac{2 + 2( {2)}^{2} - 4(5) }{4(5) + 2(2) + 1} \\ = \frac{2 + 8 - 20}{20 + 4 + 1} \\ = \frac{ - 10}{25} \\ = \frac{ - 2}{5} = \frac{ - b}{a} \: of \: new \: polynomial \\ \frac{ \beta }{2 \alpha + 1 } \times \frac{ \alpha }{2 \beta + 1} = \frac{ \alpha \beta }{4 \alpha \beta + 2( \alpha + \beta) + 1 } \\ = \frac{5}{4 \times 5 + 2(2) + 1} \\ = \frac{5}{25} \\ = \frac{1}{5} = \frac{c}{a} \: of \: new \: polynomial \\ new \: polynomial \: is \\ {x}^{2} - ( \frac{ - b}{a} )x + \frac{c}{a} \\ {x}^{2} - ( \frac{ - 2}{5} )x + \frac{1}{5} \\ {x}^{2} + \frac{2}{5}x + \frac{1}{5} \\ 5 {x}^{2} + 2x + 1 \: \: \: \: ans$