Question

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

• A. $x^2 + 10x + 34 = 0$
• B. $x^2 - 10x + 34 = 0$
• C. $x^2 - 10x - 34 = 0$
• D. $x^2 + 10x - 34 = 0$

Answer 1

Answer: (B)

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