Question

If alpha and beta are the zeroes of the equation 6a^2+x-2=0 then find alpha/beta+beta/alpha?

Answer 1

$a {x}^{2} + bx + c = 0 \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ \alpha \beta = \frac{c}{a} \\ \\ now \: here \\ \\ \bold{ \large{6 {x}^{2} + x - 2 = 0}} \\ \\ \bold{ \large{a = 6}} \\ \\ \bold{ \large{b = 1}} \\ \\ \bold{ \large{c = - 2}} \\ \\ \bold{ \large{ \alpha + \beta = - \frac{1}{6} }} \\ \\ \bold{ \large{ \alpha \beta = \frac{ - 2}{6} = - \frac{1}{3} }} \\ \\ \bold{ \large{now}} \\ \\ \bold{ \large{ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } }} \\ \\ \bold{ \large{ = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } }} \\ \\ \bold{ \large{ \frac{ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 2 \alpha \beta }{ \alpha \beta } }} \\ \\ \bold{ \large{ = \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } }} \\ \\ \bold{ \large{ = \frac{ \frac{1}{36} + \frac{1}{3} }{ - \frac{1}{3} } }} \\ \\ \bold{ \large{ = - \frac{13}{12} }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large{\color{violet}{\boxed{ \mathfrak{hope \: it \: helps}}}}$