Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+3x+1=0$, then the value of ${\left(\frac{\alpha }{1+\beta }\right)}^2+{\left(\frac{\beta }{\alpha +1}\right)}^2$ is equal to

• A. $18$
• B. $19$
• C. $20$
• D. $21$

Answer 1

The correct option is A $18$

$x^2+3x+1=0$
$\alpha +\beta =-3;\alpha \beta =1$
Also, ${\alpha }^2+3\alpha +1=0$
and ${\beta }^2+3\beta +1=0$
$\Rightarrow {\alpha }^2=-(3\alpha +1)$
and ${\beta }^2=-(3\beta +1)$

Now, ${\left(\frac{\alpha }{1+\beta }\right)}^2+{\left(\frac{\beta }{\alpha +1}\right)}^2$
$=\frac{{\alpha }^2}{(1+\beta {)}^2}+\frac{{\beta }^2}{(\alpha +1{)}^2}$
$=\frac{{\alpha }^2}{1+2\beta +{\beta }^2}+\frac{{\beta }^2}{1+2\alpha +{\alpha }^2}$
$=\left(\frac{-(1+3\alpha )}{-\beta }\right)+\left(\frac{-(1+3\beta )}{-\alpha }\right)$
$=\frac{1+3\alpha }{\beta }+\frac{1+3\beta }{\alpha }$
$=\frac{\alpha (1+3\alpha )+\beta (1+3\beta )}{\alpha \beta }$
$=3({\alpha }^2+{\beta }^2)+(\alpha +\beta )[\because \alpha \beta =1]$
$=3\left[\right(\alpha +\beta {)}^2-2\alpha \beta ]+(\alpha +\beta )$
$=3[9-2]+(-3)=21-3=18$