Question

If $\alpha and\beta$ are the zeroes of the quadratic polynomial $2S^2-8S+4$, then the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+2(\frac{1}{\alpha }+\frac{1}{\beta })+3\alpha \beta$ is ___.

• A. 8
• B. 4
• C. 16
• D. 12

Answer 1

The correct option is C 16

Given, $f\left(S\right)=2S^2-8S+4$.
Let $\alpha and\beta$ be its zeroes.
$\therefore \alpha +\beta =\frac{8}{2}=4and\alpha \beta =\frac{4}{2}=2$
Now,
$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+2(\frac{1}{\alpha }+\frac{1}{\beta })+3\alpha \beta$

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

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

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

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

= 6 + 4 + 6 = 16