If - - are the roots of the equation x2-x-1-0- then -2009-2009-

The correct option is B 1x^2 x +1 = 0 ⇒ x=1±√(1 4)/2 x= 1±√(3)i/2 α=1/2+ i√(3)/2= ω^2 β= 1/2 i√(3)/2= ω α^2009+β^2009=( ω^2)^2009+( ω)^2009= ω^2 ω=1 ...

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Standard XIII

Mathematics

Quadratic Formula for Finding Roots

if α, β are t...

Question

if $α, β$ are the roots of the equation $x^{2} - x + 1 = 0$ , then $α^{2009} + β^{2009}$ =

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Solution

The correct option is B 1
$x^{2} - x + 1 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 - 4}}{2} x = \frac{1 \pm \sqrt{3} i}{2} α = \frac{1}{2} + i \frac{\sqrt{3}}{2} = - ω^{2} β = \frac{1}{2} - \frac{i \sqrt{3}}{2} = - ω α^{2009} + β^{2009} = (- ω^{2})^{2009} + (- ω)^{2009} = - ω^{2} - ω = 1$

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if α,β are the roots of the equation x2−x+1=0, then α2009+β2009 =

The correct option is B 1x2−x+1=0⇒x=1±√1−42x=1±√3i2α=12+i√32=−ω2β=12−i√32=−ωα2009+β2009=(−ω2)2009+(−ω)2009=−ω2−ω=1

if $α, β$ are the roots of the equation $x^{2} - x + 1 = 0$ , then $α^{2009} + β^{2009}$ =

The correct option is B 1
$x^{2} - x + 1 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 - 4}}{2} x = \frac{1 \pm \sqrt{3} i}{2} α = \frac{1}{2} + i \frac{\sqrt{3}}{2} = - ω^{2} β = \frac{1}{2} - \frac{i \sqrt{3}}{2} = - ω α^{2009} + β^{2009} = (- ω^{2})^{2009} + (- ω)^{2009} = - ω^{2} - ω = 1$