Question

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

• A. 7
• B. 9
• C. 16
• D. 17

Answer 1

Answer: (A)

${\alpha }^2+3\alpha +1=0\Rightarrow (\alpha +3)=\frac{-1}{\alpha }$ and $\alpha +\beta =-3$ ${\beta }^2+3\beta +1=0\Rightarrow (\beta +3)=\frac{-1}{\beta }$ and $\alpha \beta =1$
So, ${\left(\frac{{\alpha }^2}{\frac{-1}{\beta }}\right)}^2+{\left(\frac{{\beta }^2}{\frac{-1}{\alpha }}\right)}^2={\alpha }^2{\beta }^2\left({\alpha }^2+{\beta }^2\right)=1\left((-3{)}^2-2(1)\right)=7$