Question

10. If
$\alpha \: , \: \beta \: are \: the \: zeroes \: of \: the \: \\ polynomial \: f(x) = x {}^{2} - 3x + 6, \\ then \: find \: te \: value \: of \: \frac{1}{ \alpha } + \frac{1}{ \beta } + \\ \alpha {}^{2} + \beta {}^{2} - 2 \alpha \beta$



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Answer 1

Solution:-

given by:-
$f(x) = x {}^{2} - 3x + 6,$
$\frac{1}{ \alpha } + \frac{1}{ \beta } + \\ \alpha {}^{2} + \beta {}^{2} - 2 \alpha \beta \\ we \: know \: that \\ \: \alpha + \beta = \frac{ - b}{a} = \\ \frac{3}{1} = 3 \\ \alpha \beta = \frac{c}{a} = \\ = \frac{6}{1} = 6 \\ \frac{1}{ \alpha } + \frac{1}{ \beta } + \\ \alpha {}^{2} + \beta {}^{2} - 2 \alpha \beta \\ \frac{\alpha +\beta }{\alpha \beta} + {(\alpha +\beta)}^{2} - 4\alpha \beta \\ \frac{3}{6} + {3}^{2} - 4 \times 6 \\ \frac{1}{2} + 9 - 24 \\ \frac{1 + 18 - 48}{2} \\ \frac{29}{2} ans$
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Answer 2

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