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EAMCET Engineering 2009 Solved Paper

Section: Mathematics

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Question : 99 of 160

Marks: +1, -0

If

\alpha

\beta

are the roots of

x^2-2 x+4=0

, then the value of

\alpha^6+\beta^6

32
64
128
256

Solution:

Given,

\alpha, \beta

are the roots of

x^2-2 x+4=0

\;\therefore\;\; \alpha+\beta=2

. . . (i)

\;\;\text{ and }\; \;\; \alpha\beta=4

. . . (ii)

\;\;\text{ Now, }\; \;\; \alpha-\beta=\sqrt{(\alpha+\beta)^2-4\alpha\beta}

\;=\sqrt{4-4\times 4}=\sqrt{-12}

\;\Rightarrow \;\; \alpha-\beta=2\sqrt{3}i

. . . (iii)

On solving Eqs. (i) and (ii), we get

\;\alpha=\frac{2+2\sqrt{3}i}{2}=-2\left(\frac{-1-\sqrt{3}i}{2}\right)=-2\omega^2

\;\;\text{ and }\; \beta=\frac{2-2\sqrt{3}i}{2}=-2\left(\frac{-1+\sqrt{3}i}{2}\right)=-2\omega

\;\;\text{ Now, }\; \;\; \alpha^6+\beta^6=(-2\omega^2)^6+(-2\omega)^6

\;=64(\omega^3)^4+64(\omega^3)^2

\;=128 \;\; [\because \omega^3=1]

\;

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