Question

If $\alpha ,\beta$ are roots of the equation $2x^2+6x+b=0$ where $b<0$, then find least integral value of $(\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha })$.

• A. $10$
• B. $11$
• C. $12$
• D. $15$

Answer 1

The correct option is A $10$

$\alpha +\beta =-3$ and $\alpha \beta =\frac{b}{2}$

Let $z=\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{{\alpha }^3+{\beta }^3}{\alpha \beta }$

$\Rightarrow z=\frac{(\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )}{\alpha \beta }$

$=\frac{(\alpha +\beta )\left[\right(\alpha +\beta {)}^2-3\alpha \beta ]}{\alpha \beta }=\frac{-3\left(\right(9-3\frac{b}{2})}{\frac{b}{2}}=9-\frac{54}{b}$

Now since $b<0\Rightarrow z>9$

Thus, the least integral value of $\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }$ is $10$.