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Question

Given equation x2+2x+2=0,α,β are the roots of the equation then α15+β15

A
512
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B
512
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C
256
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D
256
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Solution

The correct option is C 256
x2+2x+2=0
Here, a=1,b=2,,c=2
From quadratic formula,
x=2±224×2×12×1
x=2±2i2=1±i
Therefore,
α=1+iα2=(1+i)2=2i
β=1iβ2=(1i)2=2i
Now,
α15+β15
=(α2)7α+(β2)7β
=(2i)7(1+i)+(2i)7(1i)
=(2)7(i)(1+i)+27(i)(1i)
=27i(1+i)+27i(1+i)
=27i(1+i+1+i)
=27i2i
=28(1)
=256

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