Question

please give full explanation

Answer 1

This is the answer for question. By this method you can solve similar problems

Answer 2

Answer:

8

Step-by-step explanation:

Given : $f(x)=3x^2-6x+4$

To Find : $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha \beta$

Solution:

$\alpha$ and $\beta$ are teh zeroes of the f(x)

$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha \beta$

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

Using Identity :$(a+b)^=a^2+b^2+2ab$

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

Sum of zeroes = $\alpha +\beta = \frac{-b}{a}=\frac{-(-6)}{3}=2$

Product of zeroes = $\alpha \beta =\frac{c}{a}=\frac{4}{3}$

Substitute these values in A

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

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

$8$

So,  $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2(\frac{1}{\alpha}+\frac{1}{\beta})+3\alpha \beta=8$