Question

please give the answer​

Answer 1

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Answer 2

$6 {x}^{2} + 7x - 3$

so,

$\alpha + \beta = \frac{ - 7}{6} \\ \alpha \beta = \frac{ - 3}{6}$

so, Sum of new zeros

$\frac{1}{ \alpha } + \frac{1}{ \beta } = \frac{ { \alpha}^{2} + { \beta }^{2} }{ \alpha \beta } = \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } \\ \frac{ {( \frac{ - 7}{6}) }^{2} - 2 \times \frac{ - 3}{6} }{ \frac{ - 3}{6} } = \frac{ \frac{49}{36} + 1}{ \frac{ - 3}{6} } = \frac{ \frac{85}{36} }{ \frac{ - 3}{6} } = \frac{85 \times 6}{36 \times ( - 3)} \\ \frac{ - 85}{18}$

Product of new zeros

$\frac{1}{ \alpha } \times \frac{1}{ \beta } = \frac{1}{ \alpha \beta } = \frac{1}{ \frac{ - 3}{6} } = - 2$

so, new polynomial is

${x}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )x + \frac{1}{ \alpha \beta } \\ \\ {x}^{2} + \frac{85}{18} x - 2$

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