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Question

If α,β are the roots of the equation x2+px12p2=0, where pR. Then, the minimum value of α4+β4 is

A
2
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B
2+2
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C
22
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D
22
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Solution

The correct option is B 2+2
We have,
α+β=pαβ=12p2

α4+β4=(α2+β2)22α2β2=((α+β)22αβ)22α2β2

=((p)2+1p2)212p4

=p4+12p4+2

=(p2)2+(12p2)2+2p212p22p212p2+2

=(p212p2)2+2+2

=(p212p2)2+2+22+2

minimum value is 2+2

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