NIMCET 2011 Question Paper with solutions

Question : 10 of 120

Marks: +1, -0

α, β

are the roots of the equation

x^{2}-2 x+4=0

, then the value of

α^{6}+β^{6}

Solution:

Here,

α+β=2

and

α β=4

⇒ (α+β)^{2}=4

⇒ a^{2}+β^{2}+2 α β=4

⇒ α^{2}+β^{2}=4-2⟨ 4)=-4

⇒ α^{4}+β^{4}+2 α^{2} β^{2}=16

⇒ α^{4}+β^{4}=16-2(16)=-16

Now,

α^{8}+β^{6}=(α^{2})^{3}+(β^{2})^{3}

=(α^{2}+β^{2})(α^{4}+β^{9}-α^{2} β^{2})

=-4(-16-16)=128

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