Question

alpha and ß are the zeroes of the equation 6x^2 +x-2=0 find alpha/ ß+ß/alpha​

Answer 1

$\huge{ \sf{ \green{ \mid{ \underline{ \overline{ \pink{work \: out...}}}}}}}$

$\large{ \to \: given \: f(x) = 6 {x}^{2} + x - 2 = 0 }\\ \large{if \: \alpha \: and \: \beta \: are \: zeroes \: of \: f(x) }\\ \large{ \red{ = > sum \: of \: zeroes = \frac{ - (coefficient \: of \: x)}{coefficient \: of \: {x}^{2} } }} \\ \large{ \red{= > product \: of \: zeroes = \frac{constant}{coefficient \: of \: {x}^{2} }}} \\ \large{ \to \: then \: \alpha + \beta = \frac{ - 1}{6} ...(1) }\\ \large{ \alpha \beta = \frac{ - 2}{6} ....(2)} \\ \\ \large{ \green{ \to \: find.. \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }}} \\ = \large{ \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } } \\ \large{ = \frac{( \alpha + \beta ) {}^{2} - 2 \alpha \beta }{ \alpha \beta } }\\ \large{= \frac{( \frac{ - 1}{6}) {}^{2} - 2 \times \frac{ - 2}{6} }{ \frac{ - 2}{6} } } \\ \large{ = \frac{( \frac{1}{36} + \frac{4}{6}) }{ \frac{ - 1}{3} } }\\ \large{ = \frac{ \frac{1 + 24}{36} }{ \frac{ - 1}{3} } } \\ \large{ = \frac{ \frac{25}{36} }{ \frac{ - 1}{3} } }\\ \large{= \frac{25}{36} \times - 3 }\\ \large{ \green{ = \frac{ - 25}{12} }} (\blue{answer...})$

$\large{ \bold{ \orange{ running \: in \: a \: endless \: maze..}}}$

☘☘☘☘☘☘☘☘☘☘☘☘☘☘