Consider the equation x 2 + ax+ b = 0

If has two roots (not necessarily real )

Either or

Case (1) If , then it is repeated root. Given that is also a root

So,

or

When then (a, b) = (2, 1)

Then (a, b) = (−4, 4)

Case (2) If Then

(I) and

Here (α, β) = (2, −1) or (−1, 2)

Hence (a, b) = (− (α + β), αβ)

= (−1, −2)

(II) α = β 2 − 2and β = α 2 − 2

Then α − β = β 2 − α 2 = (β − α) (β + α)

Since α ≠ β we get α + β = β 2 + α 2 – 4

α + β = (α + β) 2 − 2αβ − 4

Thus −1 = 1 − 2αβ − 4which implies

αβ = −1Therefore (a, b) = (− (α + β), αβ)

= (1, −1)

(III) α = α 2 − 2 = β 2 – 2 and α ≠ β

⇒ α = −β

Thus α = 2, β = −2

α = −1, β = 1

Therefore (a, b) = (0, −4) & (0, −1)

(IV) β = α 2 − 2 = β 2 – 2 and α ≠ β is same as (III)

Therefore, we get 6 pairs of (a, b)

Which are (2, 1), (−4, 4), (−1, −2), (1, −1) (0, −4)