Question

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

• A. $12$
• B. $23$
• C. $13$
• D. $-13$

Answer 1

The correct option is C $13$

As $\alpha ,\beta$ are the roots of equation $2x^2-6x+3=0$.

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

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

Given expression is

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

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

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

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

$=\frac{12}{3}+3\left(\frac{6}{3}\right)+3$

$=4+6+3=13$