Question

If $\alpha, \beta$ are the roots of the equation $x^{2}-4 x+8=0$, then for any $n \in N, \alpha^{2 n}+\beta^{2 n}$ equals

• A. $2^{2 n+1} \cos \frac{n \pi}{2}$
• B. $2^{3 n} \cos \frac{n \pi}{2}$
• C. $2^{3 n+1} \cos \frac{n \pi}{2}$
• D. $2^{3 n} \cos \frac{n \pi}{4}$

Answer 1

Answer: (C)

Since, $\alpha, \beta$ are the roots of the equation
$x^{2}-4 x+8=0$
$\therefore \alpha, \beta=\frac{4 \pm \sqrt{16-32}}{2}=\frac{4 \pm 4 i}{2}=2 \pm 2 i$
$\Rightarrow \alpha, \beta=2 \sqrt{2}\left(\frac{1}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}}\right)$
$=2 \sqrt{2}\left(\cos \frac{\pi}{4} \pm i \sin \frac{\pi}{4}\right)$
Now, $\alpha^{2 n}=(2 \sqrt{2})^{2 n}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{2 n}, n \in N$
$=(2 \sqrt{2})^{2 n}\left(\cos \frac{n \pi}{2}+i \sin \frac{n \pi}{2}\right)$
Similarly, $\beta^{2 n}=(2 \sqrt{2})^{2 n}\left(\cos \frac{n \pi}{2}-i \sin \frac{n \pi}{2}\right)$
$\therefore \alpha^{2 n}+\beta^{2 n}=(2 \sqrt{2})^{2 n} \cdot 2 \cos \frac{n \pi}{2}$
$=\left(2^{3 / 2}\right)^{2 n} \cdot 2 \cos \frac{n \pi}{2}$
$\Rightarrow \alpha^{2 n}+\beta^{2 n}=2^{3 n+1} \cos \frac{n \pi}{2}$