Question

If $\alpha$ and $\beta$are the roots of the equation$x^2-x+1=0$, then${\alpha }^{2009}+{\beta }^{2009}$is equal to

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Explanation for correct option:

Step:1. The given quadratic equation :

Given that $\alpha$ and $\beta$are the roots of the equation$x^2-x+1=0$.

$\Rightarrow (\alpha +\beta )=1;\alpha .\beta =1$And also

$$\begin{gathered} x^2–x+1=0 \\ \Rightarrow (x+1)(x^2–x+1)=0 \\ \Rightarrow x^3+1=0 \\ \Rightarrow x^3=-1.......\left(i\right) \end{gathered}$$

Put $\alpha ,\beta$in equation (i). we get

${\alpha }^3=-1$and ${\beta }^3=-1$

Step:2. Find the value of $\mathbf{}{\mathit{\alpha }}^{\mathbf{2009}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{\beta }}^{\mathbf{2009}}$:

Now, ${\alpha }^{2009}+{\beta }^{2009}$

$$\begin{gathered} =\left(\frac{{\alpha }^{2010}}{\alpha }\right)+\left(\frac{{\beta }^{2010}}{\beta }\right) \\ =\frac{{\left({\alpha }^3\right)}^{670}}{\alpha }+\frac{{\left({\beta }^3\right)}^{670}}{\beta } \\ =\frac{{(-1)}^{670}}{\alpha }+\frac{{(-1)}^{670}}{\beta } \\ =\frac{1}{\alpha }+\frac{1}{\beta } \\ =\frac{(\alpha +\beta )}{\alpha \beta } \\ =1 \end{gathered}$$

Hence, the correct option is C.