Question

If alpha & beta are the zeroes of quadratic polynomial p(s) = 3s^2 - 6s + 4 , find the value of
$\alpha \div \beta + \beta \div \alpha + (1 \div \alpha + 1 \div \beta ) + 3 \alpha \beta$


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Answer 1

Answer:

13/2 or 6.5.

Step-by-step explanation:

Hope u know the formulas :D

pls just don't mind the writing

d58fd33f33e82aa5370766e1b27fd2f3.pdf

Answer 2

Step-by-step explanation:

$\alpha + \beta = \frac{6}{3} = 2$

$\alpha\beta = \frac{4}{3}$

$Now,$

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} + (\frac{1}{\alpha} + \frac{1}{\beta})+3\alpha\beta$

$\frac{\alpha²+\beta²}{\alpha\beta}+(\frac{\alpha+\beta}{\alpha\beta}) + 3\alpha\beta$

$\frac{\alpha²+\beta²+\alpha+\beta}{\alpha\beta} + 3\alpha\beta$

$\frac{(\alpha+\beta)²-2\alpha\beta+(\alpha+\beta)}{\alpha\beta} + 3\alpha\beta$

$\frac{(2)² - 2(\frac{4}{3}) + 2}{\frac{4}{3}} + 3(\frac{4}{3})$

$\frac{(4 - \frac{8}{3} + 2)}{(\frac{4}{3})} + 4$

$\frac{(6 - \frac{8}{3})}{(\frac{4}{3})} + 4$

$\frac{(\frac{10}{3})}{(\frac{4}{3})} + 4$

$\frac{10}{4} + 4$

$\frac{5}{2} + \frac{8}{2}$

$\frac{13}{2}$