Question

If $\alpha$ and $\beta$ are the roots of the equation $2x^2+6x+b=0(b<0,$ then $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$ is less than

• A. $2$
• B. $-2$
• C. $18$
• D. None of these

Answer 1

Answer: (B)

$-2$

$\alpha$ and $\beta$ are roots of the equation $2x^2+6x+b=0.$

Comparing given equation with $A{x}^2+Bx+C=0$

We get, $A=2,B=6$ and $C=b$

$\Rightarrow$ $\alpha +\beta =\frac{-B}{A}=\frac{-6}{2}=-3$ ----- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{C}{A}=\frac{b}{2}$ -------- ( 2 )

$\Rightarrow$ $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$ $=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

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

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

$=\frac{(-3{)}^2}{\frac{b}{2}}-2$ [ From ( 1 ) and ( 2 ) ]

$=\frac{18}{b}-2$

Since, $b<0$

The maximum value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$ is $-2\left(approx\right)$