Question

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