Question

Let $\alpha$ and $\beta$ be two roots of the equation $x^2+2x+2=0$, then ${\alpha }^{15}+{\beta }^{15}$ is equal to :

• A. $256$
• B. $-256$
• C. $512$
• D. $-512$

Answer 1

The correct option is B $-256$

Given :
$x^2+2x+2=0$ and its roots are $\alpha$, $\beta$.
$x^2+2x+1+1=0$
$\Rightarrow (x+1{)}^2+1=0$ [Since, $a^2+2ab+b^2=(a+b{)}^2$]
$\Rightarrow (x+1{)}^2=-1$
$\Rightarrow x+1=\sqrt{-1}$
$\Rightarrow x_1=-1+i;x_2=-1-i$ [Since, $\sqrt{-1}=i$]
Assuming $\alpha =-1+i$ and $\beta =-1-i$
Squaring on both sides, we get
$\Rightarrow {\alpha }^2=(-1+i{)}^2$ and ${\beta }^2=(-1-i{)}^2$
$\Rightarrow {\alpha }^2=1+i^2-2i$ and ${\beta }^2=1+i^2+2i$
$\Rightarrow {\alpha }^2=1-1-2i$ and ${\beta }^2=1-1+2i$ [Since, $i^2=-1$]
$\therefore {\alpha }^2=-2i$ and ${\beta }^2=2i$
Now consider, ${\alpha }^{15}+{\beta }^{15}=({\alpha }^{14}\times \alpha )+({\beta }^{14}\times \beta )$
$=\left[\right({\alpha }^2{)}^7\times \alpha ]+[({\beta }^2{)}^7\times \beta )]$
$=(-2i{)}^7\times (-1+i)+(2i{)}^7\times (-1-i)$
$=-2^7{i}^7(-1+i)-2^7{i}^7(1+i)$
$=-2^7{i}^7(-1+i+1+i)$
$=-128i^7\left(2i\right)$
$=(-128\times 2)\times i^8$
$=-256(i^2{)}^4$
$=-256(-1{)}^4$
$\therefore {\alpha }^{15}+{\beta }^{15}=-256$