Question

$\alpha ,\beta$ are the solutions of $x^2+2x+2=0$ then ${\alpha }^{15}+{\beta }^{15}$ is

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

Answer 1

The correct option is C $-256$

$x^2+2x+2=0$

Here, $a=1,b=2,,c=2$

From quadratic formula,

$x=\frac{-2\pm \sqrt{2^2-4\times 2\times 1}}{2\times 1}$

$\Rightarrow x=\frac{-2\pm 2i}{2}=-1\pm i$

Therefore,

$\alpha =-1+i\Rightarrow {\alpha }^2={(-1+i)}^2=-2i$

$\beta =-1-i\Rightarrow {\beta }^2={(-1-i)}^2=2i$

Now,

${\alpha }^{15}+{\beta }^{15}$

$={\left({\alpha }^2\right)}^7\cdot \alpha +{\left({\beta }^2\right)}^7\cdot \beta$

$={(-2i)}^7(-1+i)+{\left(2i\right)}^7(-1-i)$

$={(-2)}^7(-i)(-1+i)+2^7(-i)(-1-i)$

$=2^{7}i(-1+i)+2^{7}i(1+i)$

$=2^{7}i(-1+i+1+i)$

$=2^{7}i\cdot 2i$

$=2^8\cdot (-1)$

$=-256$