Consider the equation x² + ax + b = 0

If has two roots (not necessarily real $$\alpha$$ & $$\beta$$)

Either $$\alpha$$ = $$\beta$$ or $$\alpha$$ $$\ne$$ $$\beta$$

Case (1) If $$\alpha$$ = $$\beta$$, then it is repeated root. Given that $$\alpha$$² $$-$$ 2 is also a root

So, $$\alpha$$ = $$\alpha$$² $$-$$ 2 $$\Rightarrow$$ ($$\alpha$$ + 1)($$\alpha$$ $$-$$ 2) = 0

$$\Rightarrow$$ $$\alpha$$ = $$-$$1 or $$\alpha$$ = 2

When $$\alpha$$ = $$-$$1 then (a, b) = (2, 1)

$$\alpha$$ = 2 then (a, b) = ($$-$$4, 4)

Case (2) If $$\alpha$$ $$\ne$$ $$\beta$$

Then

(I) $$\alpha$$ = $$\alpha$$² $$-$$ 2 and $$\beta$$ = $$\beta$$² $$-$$ 2

Hence, (a, b) = ($$-$$($$\alpha$$ + $$\beta$$), $$\alpha$$$$\beta$$)

($$-$$1, $$-$$2)

(II) $$\alpha$$ = $$\beta$$² $$-$$ 2 and $$\beta$$ = $$\alpha$$² $$-$$ 2

Then $$\alpha$$ $$-$$ $$\beta$$ = $$\beta$$² $$-$$ $$\alpha$$² = ($$\beta$$ $$-$$ $$\alpha$$) ($$\beta$$ + $$\alpha$$)

Since $$\alpha$$ $$\ne$$ $$\beta$$ we get $$\alpha$$ + $$\beta$$ = $$\beta$$² + $$\alpha$$² $$-$$ 4

$$\alpha$$ + $$\beta$$ = ($$\alpha$$ + $$\beta$$)² $$-$$ 2$$\alpha$$$$\beta$$ $$-$$ 4

Thus $$-$$1 = 1 $$-$$2 $$\alpha$$$$\beta$$ $$-$$ 4 which implies

$$\alpha$$$$\beta$$ = $$-$$1 Therefore (a, b) = ($$-$$($$\alpha$$ + $$\beta$$), $$\alpha$$$$\beta$$)

= (1, $$-$$1)

(III) $$\alpha$$ = $$\alpha$$² $$-$$ 2 = $$\beta$$² $$-$$ 2 and $$\alpha$$ $$\ne$$ $$\beta$$

$$\Rightarrow$$ $$\alpha$$ = $$-$$ $$\beta$$

Thus $$\alpha$$ = 2, $$\beta$$ = $$-$$2

$$\alpha$$ = $$-$$1, $$\beta$$ = 1

Therefore (a, b) = (0, $$-$$4) & (0, 1)

(IV) $$\beta$$ = $$\alpha$$² $$-$$ 2 = $$\beta$$² $$-$$ 2 and $$\alpha$$ $$\ne$$ $$\beta$$ is same as (III) Therefore we get 6 pairs of (a, b)

Which are (2, 1), ($$-$$4, 4), ($$-$$1, $$-$$2), (1, $$-$$1), (0, $$-$$4)

Option (a)