Question

if alpha and bita are the zeroes of the polynomial 3x^2+2x-6 , then find the value of alpha/bita + bita/alpha​

Answer 1

Step-by-step explanation:

Given quadratic polynomial is:

$3 {x}^{2} + 2x - 6 \\ equating \: it \: with \\ a {x}^{2} + bx + c \: we \: find : \\ a = 3 \\ b = 2 \\ c = - 6 \\ \because \: \alpha \: and \: \beta \: are \: zeros \: of \: polynomial \\ hence \\ sum \: of \: zeros \\ \alpha + \beta = - \frac{b}{a} = - \frac{2}{3} \\ product \: of \: zeros \\ \alpha \beta = \frac{c}{a} = \frac{ - 6}{3} = - 2 \\ now \: we \: need \: to \: find : \\ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } \\ thus \\ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } = \frac{( \alpha + \beta )^{2} - 2 \alpha \beta }{ \alpha \beta } \\ = \frac{ ( - \frac{2}{3})^{2} - 2( - 2) }{ - 2} \\ = \frac{ \frac{4}{9} + 4 }{ - 2} = \frac{ \frac{4 + 9 \times 4}{9} }{2} \\ = \frac{ \frac{4 + 36}{9} }{ - 2} \\ = \frac{ 40}{ - 2 \times 9} \\ = - \frac{20}{9} \\ \therefore \: \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } = - \frac{20}{9}$