Question

Maths lovers ......
One more challenge for you !

Answer 1

$Answer \: \\ \\ Given \: \: Quadratic \: \: Polynomial \: is \: \\ \\ p(x) = ax {}^{2} + bx + c \\ \\ let \: its \: two \: zeroes \: be \: \: \: \alpha \: \: \: and \: \: \beta \\ \\ \alpha + \beta = \frac{ - b}{a} \: \: \: and \: \: \: \: \alpha \beta = \frac{c}{a} \\ \\ RECIPROCAl \: \: Of \: \: these \: zeroes \: are \\ \\ \frac{1}{ \alpha } \: \: \: \: and \: \: \: \: \frac{1}{ \beta } \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \alpha + \beta }{ \alpha \beta } \\ \\ \frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ \frac{ - b}{a} }{ \frac{c}{a} } \\ \\ \frac{1 }{ \alpha } + \frac{1}{ \beta } = \frac{ - b}{c} \: \: \: \: \: \: \: \: and \: \: \: \: \frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{ \frac{c}{a} } = \frac{a}{c} \\ \\ let \: the \: Required \: Quadratic \: polynomial \: be \: \: \\ f(x) \\ \\ f(x) = x {}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )x + \frac{1}{ \alpha \beta } \\ \\ f(x) = x {}^{2} - ( \frac{ - b}{c} )x + \frac{a}{c} \\ \\ f(x) = x {}^{2} + ( \frac{b}{c} )x + \frac{a}{c} \\ \\ therefore \: \: required \: quadratic \: polynomial \: is \: \: \\ \\ x {}^{2} + ( \frac{b}{c} )x + \frac{a}{c} \\ and \: \: required \: quadratic \: equation \: is \: \: \\ \\ x {}^{2} + ( \frac{b}{c} )x + \frac{a}{c} = 0 \\ \\ \\ Note \: \\ \\ for \: a \: general \: quadratic \: polynomial \: \\ say \: \: g(x) = px {}^{2} + qx + v \\ \\ \alpha + \beta = \frac{ - q}{p} \: \: \: \: and \: \: \: \: \alpha \beta = \frac{v}{p}$