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Q.

If α,β,γ,δR satisfy (α+1)2+(β+1)2+(γ+1)2+(δ+1)2α+β+γ+δ=4 if biquadratic equation a0x4+a1x3+a2x2+a3x+a4=0has the roots (α+1β1),(β+1γ1),(γ+1δ1),(δ+1α1). Then the value of a2 / a0 is

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a

-4

b

6

c

3

d

4

answer is C.

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Detailed Solution

α=1,β=1,γ=1,δ=1 as (α1)2+(β1)2+(γ1)2+(δ1)2=0

The roots of given equation is equal to 1
S2=a2a0=6

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