If #alpha# and #beta# are the roots of the equation #x^2 - x +1 = 0#, then #alpha^2009 + beta^2009# is?

A) -2
B) -1
C) 1
D) 2

1 Answer
George C.
Aug 18, 2017

C) 1

Explanation:

Note that:

#x^3+1 = (x+1)(x^2-x+1)#

So since #alpha# is a root of #x^2-x+1 = 0# we have:

#alpha^3+1 = (alpha+1)(alpha^2-alpha+1) = (alpha+1)(0) = 0#

So:

#alpha^3 = -1#

Similarly:

#beta^3 = -1#

Also:

#x^2-x+1 = (x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta#

So:

#{ (alpha+beta = 1), (alphabeta=1) :}#

Note that:

#alpha^2 + beta^2 = (alpha+beta)^2-2alphabeta = 1-2 = -1#

So:

#alpha^2009+beta^2009 = alpha^(3*669+2)+beta^(3*669+2)#

#color(white)(alpha^2009+beta^2009) = (alpha^3)^669*alpha^2+(beta^3)^669*beta^2#

#color(white)(alpha^2009+beta^2009) = (-1)^669*alpha^2+(-1)^669*beta^2#

#color(white)(alpha^2009+beta^2009) = (-1)^669*(alpha^2+beta^2)#

#color(white)(alpha^2009+beta^2009) = (-1)(-1)#

#color(white)(alpha^2009+beta^2009) = 1#

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