Solution:
Given equation x2+ax+b=0
It has two roots (not necessarily real α and β )
⇒ Either α=β or α≠β
1. If α=β⇒α=α2−2⇒α=−1,2
When α=−1, then (a,b)=(2,1)
When α=2, then (a,b)=(−4,4)
II. If α≠β, then
(a) α=α2−2 and β=β2−2
Here, (α,β)=(2,−1) or (−1,2)
Hence (a,b)=(−α−β,αβ)=(−1,−2)
(b) α=β2−2 and β=α2−2
Then α−β=β2−α2=(β−α)(β+α)
∵‌‌α≠β
⇒‌‌α+β=β2+α2−4
or α+β=(α+β)2−2αβ−4
⇒‌‌−1=1−2αβ−4⇒αβ=−1
⇒‌‌(a,b)=(−α−β,αβ)=(1,−1)
(c) α=α2−2=β2−2 and α≠β⇒α=−β
Thus, α=2,β=−2
or ‌‌α=−1,β=1
∴‌‌(a,b)=(0,−4) and (0,−1)
(d) β=α2−2=β2−2 and α≠β( as in (c))
⇒ We get 6 pairs of (a,b)
They are (2,1),(−4,4),(−1,−2),(1,−1),
(0,−4), and (0,−1).
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