If alphaα and betaβ are the roots of the equation x^2 - x +1 = 0x2−x+1=0, then alpha^2009 + beta^2009α2009+β2009 is?
A) -2
B) -1
C) 1
D) 2
A) -2
B) -1
C) 1
D) 2
1 Answer
C) 1
Explanation:
Note that:
x^3+1 = (x+1)(x^2-x+1)x3+1=(x+1)(x2−x+1)
So since
alpha^3+1 = (alpha+1)(alpha^2-alpha+1) = (alpha+1)(0) = 0α3+1=(α+1)(α2−α+1)=(α+1)(0)=0
So:
alpha^3 = -1α3=−1
Similarly:
beta^3 = -1β3=−1
Also:
x^2-x+1 = (x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabetax2−x+1=(x−α)(x−β)=x2−(α+β)x+αβ
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
