Question

If $\alpha ,\beta$ be the roots of the equation $3x^2-6x+4=0$ then the value of $(\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha })+(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })+2(\frac{1}{\alpha }+\frac{1}{\beta })+3a\beta$ is

Answer 1

Given,

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

$=\frac{(\alpha +\beta )\left[\right(\alpha +\beta {)}^2-3\alpha \beta ]+(\alpha +\beta {)}^2-2\alpha \beta +2(\alpha +\beta )+3{\alpha }^2{\beta }^2}{\alpha \beta }$

Given,

$3x^2-6x+4=0$

$\Rightarrow \alpha +\beta =2,\alpha \beta =\frac{4}{3}$

$=\frac{\left(2\right)\left[\right(2{)}^2-3\times \frac{4}{3}]+(2{)}^2-2\times \frac{4}{3}+2(2)+3{\left(\frac{4}{3}\right)}^2}{\frac{4}{3}}$

$=\frac{0+4-\frac{8}{3}+4+\frac{16}{3}}{\frac{4}{3}}$

$=\frac{8}{4}=2$