Question

if alpha and beta are the roots of the equation 3x² minus 11x plus 6≈0 find the
1×beta
alpha² plus beta²
alpha minus beta

Answer 1

Question Solved in Paper Plz see the pic for Answer

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Answer 2

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

${\bold{\therefore \beta\alpha^{2}+\beta^{2}\alpha-\alpha=\frac{13}{3}}}$

${\bold{\therefore\beta\alpha^{2}+\beta^{2}\alpha-\alpha=6}}$

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

$\underline \bold{Given : } \\ \implies \alpha \: and \: \beta \in \: (3 {x}^{2} - 11x + 6 = 0) \\ \\ \underline \bold{To \: Find : } \\ \implies \beta { \alpha }^{2} - { \beta }^{2} \alpha - \beta = ?$

• According to given question :

$\bold{Using \: middle \: term \: spliting : } \\ \implies {3x}^{2} - 11x + 6 = 0 \\ \\ \implies {3x}^{2} - 9x - 2x + 6 = 0 \\ \\ \implies 3x(x - 3) - 2(x - 3) = 0 \\ \\ \implies (3x - 2)(x - 3) = 0 \\ \\ \bold{Case \: I} \\ \bold{if \: \alpha = \frac{ 2}{3} \: and \: \beta = 3} \\ \\ \implies \beta { \alpha }^{2} + { \beta }^{2} \alpha - \beta \\ \\ \implies 3 \times( \frac{2}{3})^{2} + {3}^{2} \times \frac{2}{3} - 3 \\ \\ \implies \frac{4}{3} + 6 - 3 \\ \\ \implies \frac{4}{3} + 3 \\ \\ \implies \frac{4 + 9}{3} \\ \\ \bold{\implies \frac{13}{3} } \\ \\ \bold{ Case \: II} \\ \bold{if \: \alpha = 3 \: and \: \beta = \frac{2}{3} } \\ \\ \implies \beta { \alpha }^{2} + { \beta }^{2} \alpha - \beta \\ \\ \implies \frac{2}{3} \times {3}^{2} + (\frac{2}{3})^{2} \times 3 - \frac{2}{3} \\ \\ \implies 6 + \frac{2}{3} - \frac{2}{3} \\ \\ \bold{\implies 6}$