Question

Challenger question for maths lovers

Answer 1

$Answer \: \\ \\ Given \: \: Quadratic \: \: polynomials \: \: is \: \\ \\ f(x) = x {}^{2} - 3x - 2 \\ having \: \: \alpha \: and \: \beta \: its \: two \: roots \\ \\ \alpha + \beta = \frac{ 3}{1} \: \: \: and \: \: \: \alpha \beta = \frac{ - 2}{ 1} \\ \\ \alpha + \beta = 3 \: \: \: \: and \: \: \: \alpha \beta = - 2 \\ \\ let \: the \: required \: quadratic \: polynomial \: be \: p(x) \\ \\ p(x) = x {}^{2} - ( \frac{1}{2 \alpha + \beta } + \frac{1}{2 \beta + \alpha })x + ( \frac{1}{2 \alpha + \beta } \times \frac{1}{2 \beta + \alpha } ) \\ \\ p(x) = x {}^{2} - ( \frac{2 \beta + \alpha + 2 \alpha + \beta) }{(2 \alpha + \beta )(2 \beta + \alpha )}) x + \frac{1}{(2 \alpha + \beta )(2 \beta + \alpha )} \\ \\ p(x) = x {}^{2} - ( \frac{3 \alpha + 3 \beta }{4 \alpha \beta + 2 \alpha {}^{2} + 2 \beta {}^{2} + \alpha \beta } )x + \frac{1}{(5 \alpha \beta + 2( \alpha {}^{2} + \beta {}^{2} ))} \\ \\ p(x) = x {}^{2} - ( \frac{3( \alpha + \beta )}{5 \alpha \beta + 2( \alpha {}^{2} + \beta {}^{2}) } ) x+ \frac{1}{5 \alpha \beta + 2( \alpha { }^{2} + \beta {}^{2} ) } \\ \\ p(x) = x {}^{2} - (\frac{3(3)}{5( - 2) + 2((3) {}^{2} - 2( - 2))}) x + \frac{1}{5( - 2) + 2((3) {}^{2} - 2( - 2))} \\ \\ p(x) = x {}^{2} - (\frac{ 9}{ 16} )x + \frac{1}{ 16} \\ \\ p(x) = x {}^{2} - ( \frac{9}{16} )x + \frac{1}{16} \\ \\ p(x) =(16x {}^{2} - 9x + 1) \frac{1}{16} \\ \\ so \: the \: required \: quadratic \: polynomial \: is \: \\ \\ p(x) = (16x {}^{2} - 9x + 1) \frac{1}{16} \\ \\ and \: the \: required \: equation \: is \\ \\ 16x {}^{2} - 9x + 1 = 0 \\ \\ NOTE \: \\ \\ ( \alpha {}^{2} + \beta {}^{2} ) = ( \alpha + \beta ) {}^{2} - 2 \alpha \beta$