Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $3t^2-6t+4$, find the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+2\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)+3\alpha \beta$.

• A. $5$
• B. $8$
• C. $\frac{10}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $8$

$\alpha +\beta$ is $2$ and $\alpha \beta$ is $4/3$ using same in the below also..

as simplified form of the question is $\left[\right((\alpha +\beta {)}^2+2(\alpha +\beta )-(2\alpha \beta \left)\right)\left]/\right(\alpha \beta )+3\alpha \beta$

so using the above roots in it we get the answer as 8.

Hence option B is correct.