Question

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-2 x+4=0$, then $\alpha^{9}+\beta^{9}$ is equal to

• A. $-2^{8}$
• B. $2^{9}$
• C. $-2^{10}$
• D. $2^{10}$

Answer 1

Answer: (C)

Given quadratic equation is $x^{2}-2 x+4=0$
whose roots are $\alpha$ and $\beta$.
$\therefore \alpha+\beta=2$ and $\alpha \beta=4$ ...(i)
Now,
$\alpha^{9}+\beta^{9}=\left(\alpha^{3}\right)^{3}+\left(\beta^{3}\right)^{3}$
$=\left(\alpha^{3}+\beta^{3}\right)\left(\alpha^{6}+\beta^{6}-\alpha^{3} \beta^{3}\right)$
$=(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right)\left[\left(\alpha^{2}\right)^{3}+\left(\beta^{2}\right)^{3}-\alpha^{3} \beta^{3}\right]$
$=(\alpha+\beta)\left[(\alpha+\beta)^{2}-3 \alpha \beta\right]$
$\left[\left(\alpha^{2}+\beta^{2}\right)\left(\alpha^{4}+\beta^{4}-\alpha^{2} \beta^{2}\right)-\alpha^{3} \beta^{3}\right]$
$=(\alpha+\beta)\left[(\alpha+\beta)^{2}-3 \alpha \beta\right]\left[\left\{(\alpha+\beta)^{2}-2 \alpha \beta\right\}\right.$
$\left.\left\{\left(\alpha^{2}+\beta^{2}\right)^{2}-3 \alpha^{2} \beta^{2}\right\}-\alpha^{3} \beta^{3}\right]$
$=(\alpha+\beta)\left[(\alpha+\beta)^{2}-3 \alpha \beta\right]\left[\left\{(\alpha+\beta)^{2}-2 \alpha \beta\right\}\right.$
$\left.\left[\left\{(\alpha+\beta)^{2}-(2 \alpha \beta)\right]^{2}-3 \alpha^{2} \beta^{2}\right\}-\alpha^{3} \beta^{3}\right]$
$= 2[4-12]\left[\{4-8\}\left\{(4-8)^{2}-48\right\}-64\right]$ [from Eq. (i)]
$= 2(-8)\{(-4)(-32)(-64)\}$
$= 2(-8)(128-64)$
$= 2(-8)(64)=-2^{10}$