If $\alpha$ and $\beta$ are the roots of the equation $x^2-4 x+5=0$, then the quadratic equation whose roots are $\alpha^2+\beta$ and $\alpha+\beta^2$ is

Question

Aicharya · Accepted Answer

Since, $\alpha$ and $\beta$ are roots of the quadratic equation $ x^2-4 x+5=0 $

Now, $\left(\alpha^2+\beta\right)+\left(\alpha+\beta^2\right)=\left(\alpha^2+\beta^2\right)+(\alpha+\beta)$ $ \begin{aligned} & =(\alpha+\beta)^2-2 \alpha \beta+(\alpha+\beta) \\ & =16-10+4=10 \end{aligned} $ and $\left(\alpha^2+\beta\right)$ $ \begin{aligned} & \left(\alpha+\beta^2\right)=\alpha^3+\alpha^2 \beta^2+\beta \alpha+\beta^3 \\ & =\alpha^3+\beta^3+\alpha \beta(\alpha \beta+1) \\ & =(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)+\alpha \beta(\alpha \beta+1) \\ & =(\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]+\alpha \beta(\alpha \beta+1) \\ & =4[16-15]+5(5+1) \\ & =4+30=34 \end{aligned} $ So, the quadratic equation whose roots are $\left(\alpha^2+\beta\right)$ and $\left(\alpha+\beta^2\right)$ is $ \begin{aligned} x^2-\left(\alpha^2+\beta+\alpha+\beta^2\right) x+\left(\alpha^2+\beta\right)\left(\alpha+\beta^2\right) & =0 \\ \Rightarrow \quad x^2-10 x+34 & =0 \end{aligned} $

If $\alpha$ and $\beta$ are the roots of the equation $x^2-4 x+5=0$, then the quadratic equation whose roots…

Solution