Question

If $\alpha$and $\beta$ are the roots of the equation$x^2-2x+4=0$, then the value ${\alpha }^n+{\beta }^n$of will be

• A. $i{\left(2\right)}^{n+1}\sin (n\pi /3)$
• B. ${\left(2\right)}^{n+1}\cos (n\pi /3)$
• C. $i{\left(2\right)}^{n-1}\sin (n\pi /3)$
• D. ${\left(2\right)}^{n-1}\cos (n\pi /3)$

Answer 1

The correct option is B

${\left(2\right)}^{n+1}\cos (n\pi /3)$

Explanation for the correct option:

Step1. Find the value of $\mathit{\alpha }$and $\mathit{\beta }$ :

Given that $\alpha$and $\beta$ are the roots of the equation$x^2-2x+4=0$

$\Rightarrow (\alpha +\beta )=2........\left(i\right)$ and $\alpha .\beta =4$

$$\begin{gathered} \Rightarrow (\alpha –\beta )=\sqrt{{(\alpha +\beta )}^2–4\alpha \beta } \\ \Rightarrow (\alpha –\beta )=\sqrt{4-16} \\ \Rightarrow (\alpha –\beta )=2\sqrt{3}i.........\left(ii\right) \end{gathered}$$

Adding equations (i) and (ii). we get

$$\begin{gathered} 2\alpha =2+2\sqrt{3}i \\ \alpha =2\left[\frac{1}{2}+\frac{\sqrt{3}}{2}i\right] \\ \alpha =2[\cos (\pi /3)+i\sin (\pi /3\left)\right] \end{gathered}$$

Putting the value of $\alpha$in equation (i). we get

$$\begin{gathered} \Rightarrow \beta =\frac{1}{2}[2–2\sqrt{3}i] \\ \Rightarrow \beta =2[\cos (\pi /3)+i\sin (\pi /3\left)\right] \end{gathered}$$

Step2. Find the value of ${\mathit{\alpha }}^{\mathbf{n}}\mathbf{+}{\mathit{\beta }}^{\mathbf{n}}$:

$$\begin{gathered} {\alpha }^n+{\beta }^n={[2\cos (\pi /3)+i\sin (\pi /3\left)\right]}^n+{[2\cos (\pi /3)+i\sin (\pi /3\left)\right]}^n \\ \Rightarrow {\alpha }^n+{\beta }^n=2^n[2\cos (n\pi /3\left)\right] \\ \Rightarrow {\alpha }^n+{\beta }^n=2^{n+1}\cos (n\pi /3) \end{gathered}$$

Hence, the correct option is B.