Question

If \[\alpha \] and $\beta $ are the roots of the equation ${x^2} - x + 1 = 0$ , then ${\alpha ^{2009}} + {\beta ^{2009}}$ =
A $ - 1$
B $1$
C $2$
D $ - 2$

Answer 1

Hint: In this find the \[\alpha \] and $\beta $ by using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ hence you will find that it will in complex form and it is the cube root of unity . Hence we know that ${\omega ^{3x}} = 1$ and $1 + \omega + {\omega ^2} = 0$ where $\omega $ is the cube root of unity from using this we can find out the value of ${\alpha ^{2009}} + {\beta ^{2009}}$.

Complete step-by-step answer:
In this question first we have to find out the roots of equation ${x^2} - x + 1 = 0$
by using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for the equation $a{x^2} + bx + c = 0$
For the equation ${x^2} - x + 1 = 0$ $a = 1,b = - 1,c = 1$
$x = \dfrac{{1 \pm \sqrt {1 - 4} }}{2}$
$x = \dfrac{{1 \pm \sqrt { - 3} }}{2}$
$x = \dfrac{{1 + i\sqrt 3 }}{2},x = \dfrac{{1 - i\sqrt 3 }}{2}$
$\alpha = \dfrac{{1 + i\sqrt 3 }}{2}$$\beta = \dfrac{{1 - i\sqrt 3 }}{2}$
As we know that the cube root of ${\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}$$\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2}$
Hence ,
$ - {\omega ^2} = \alpha = \dfrac{{1 + i\sqrt 3 }}{2}$ and $ - \omega = \beta = \dfrac{{1 - i\sqrt 3 }}{2}$
Now,
${\alpha ^{2009}} + {\beta ^{2009}}$
${( - {\omega ^2})^{2009}} + {( - \omega )^{2009}}$
Hence we know that ${\omega ^{3x}} = 1$
${( - \omega )^{4018}} + {( - \omega )^{2009}}$
$ - \omega .{\omega ^{4017}} - {\omega ^2}.{\omega ^{2007}}$
$ - \omega .{\omega ^{3 \times 1339}} - {\omega ^2}.{\omega ^{3 \times 669}}$
${\omega ^{3 \times 1339}} = {\omega ^{3 \times 669}} = 1$ by the property ${\omega ^{3x}} = 1$
Therefore ,
$ - \omega - {\omega ^2}$
$ - (\omega + {\omega ^2})$
As we know for the cube root of unity $1 + \omega + {\omega ^2} = 0$ therefore $ - ( - 1)$

Hence the correct answer is option B.

Note: In general, if n be a positive integer then,
$\
  {\omega ^{3n}} = {({\omega ^3})^n} = {1^n} = 1 \\
  {\omega ^{3n + 1}} = {\omega ^{3n}}.\omega = 1.\omega = \omega \\
  {\omega ^{3n + 2}} = {\omega ^{3n}}.{\omega ^2} = 1.{\omega ^2} = {\omega ^2} \\
\ $
Always remember the property of cube roots of unity as $1 + \omega + {\omega ^2} = 0$ , ${\omega ^{3x}} = 1$ and $1.\omega .{\omega ^2} = 1$ .
In this type of question when $\alpha = \dfrac{{1 + i\sqrt 3 }}{2}$$\beta = \dfrac{{1 - i\sqrt 3 }}{2}$ then we have to know that it is the cube roots of unity then further solve it according to that .