Question

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

• A. $\frac{-15}{2}$
• B. $\frac{6}{5}$
• C. $\frac{-5}{2}$
• D. $-5$

Answer 1

The correct option is D $-5$

$\Rightarrow$ $\alpha$ and $\beta$ are roots of the given equation $x^2+3x-2=0$

Here, $a=1,b=3,c=-2$

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{-2}{1}=-2$ ---- ( 1 )

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-3}{1}=-3$ ----- ( 2 )

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

$\Rightarrow$ $(-3{)}^2=\alpha {\beta }^2+2(-2)$ [ Using ( 1 ) and ( 2 ) ]

$\Rightarrow$ $9={\alpha }^2+{\beta }^2-4$

$\therefore$ ${\alpha }^2+{\beta }^2=13$ ---- ( 3 )

$\frac{\alpha +1}{\beta }+\frac{\beta +1}{\alpha }=\frac{{\alpha }^2+\alpha +{\beta }^2+\beta }{\alpha \beta }$

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

$=\frac{13-3}{-2}$ [ From ( 1 ), ( 2 ) and ( 3 ) ]

$=\frac{-10}{2}$

$=-5$

$\therefore$ $\frac{\alpha +1}{\beta }+\frac{\beta +1}{\alpha }=-5$