Question

If $ɑ,\beta$ are the roots of the equation $x^2-x+1=0$ then ${ɑ}^{2009}+{\beta }^{2009}=$

• A. $-1$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Step 1. Find the value of ${ɑ}^{2009}+{\beta }^{2009}$:

As we know,

$\begin{array}{rcl}{x}^3+1& =& (x+1)(x^2–x+1)\\ & =& 0\end{array}$

$\Rightarrow$ $x^3=–1$

Step 2. $ɑ,\beta$ are the roots of the equation $x^2-x+1=0$, then

$(ɑ+\beta )=1,\left(ɑ\beta \right)=1$

$\because x^3+1=0$

$\Rightarrow$${ɑ}^3=–1,{\beta }^3=–1$

$\begin{array}{rcl}\therefore {ɑ}^{2009}+{\beta }^{2009}& =& \frac{ɑ2010}{ɑ}+\frac{\beta 2010}{\beta }\\ & =& \frac{{\left({ɑ}^3\right)}^{670}}{ɑ}+\frac{{\left({\beta }^3\right)}^{670}}{\beta }\\ & =& \frac{1}{ɑ}+\frac{1}{\beta }\\ & =& \frac{\left(\alpha +\beta \right)}{\alpha \beta }\\ & =& \frac{1}{1}\\ & =& \mathbf{1}\end{array}$

Hence, Option ‘B’ is Correct.