Question

If alpha and beta are the zeros of the quadratic polynomial f (x)=6x² + x - 2 then find the value of

•1/α+1/β −αβ

•αβ² + α²β


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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{1}{\alpha}+\frac{1}{\beta}-\alpha\beta=\frac{5}{6}}}}$

$\green{\tt{\therefore{\alpha\beta^{2}+\alpha^{2}\beta=\frac{1}{18}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies f(x) = {6x}^{2} + x - 2 \\ \\ \tt: \implies \alpha \: and \: \beta \: are \: zeroes \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta = ? \\ \\ \tt: \implies \alpha { \beta }^{2} + { \alpha }^{2} \beta =?$

• According to given question :

$\tt \circ \: f(x) = {6x}^{2} + x - 2 \\ \\ \tt \circ \: a = 6 \: \: \: \: \: \: b = 1 \: \: \: \: \: \: c = - 2 \\ \\ \bold{For \: sum \: of \: zeroes} \\ \tt: \implies sum \: of \: zeroes = \frac{ - b}{a} \\ \\ \tt: \implies \alpha + \beta = \frac{ - 1}{6} - - - - - (1) \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies product \: of \: zeroes = \frac{c}{a} \\ \\ \tt: \implies \alpha \beta = \frac{ - 2}{6} \\ \\ \tt: \implies \alpha \beta = \frac{ - 1}{3} - - - - - (2) \\ \\ \bold{For \: finding \: value } \\ \tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta$

$\tt: \implies \frac{ \beta + \alpha }{ \alpha \beta } - \alpha \beta \\ \\ \tt: \implies \frac{ \frac{ - 1}{6} }{ \frac{ - 1}{3} } - (\frac{ - 1}{3} ) \\ \\ \tt: \implies \frac{3}{6} + \frac{1}{3} \\ \\ \tt: \implies \frac{3 + 2}{6} \\ \\ \green{\tt: \implies \frac{1}{ \alpha } + \frac{1}{ \beta } - \alpha \beta = \frac{5}{6} } \\ \\ \bold{For \: another \: value } \\ \tt: \implies { \alpha \beta }^{2} + { \alpha }^{2} \beta \\ \\ \tt: \implies \alpha \beta ( \beta + \alpha ) \\ \\ \tt: \implies \frac{ - 1}{3} \times \frac{ - 1}{6} \\ \\ \green{\tt: \implies \alpha { \beta }^{2} + { \alpha }^{2} \beta = \frac{1}{18} }$