Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+3 x+1=0$ 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, $\left(\frac{\alpha^2}{\frac{-1}{\beta}}\right)^2+\left(\frac{\beta^2}{\frac{-1}{\alpha}}\right)^2=\alpha^2 \beta^2\left(\alpha^2+\beta^2\right)=1\left((-3)^2-2(1)\right)=7$