Let abcd be four integers such that ad is odd and bc is even, then ax3bx2 + cx + d = 0       ... (1)      



Correct option is

At least one irrational root

Putting ax = y (1) can be written as

                   y3 + by2 + acy + a2d = 0         ... (2)

If (1) has all rational roots then (2) has all integral roots. If α, β, γ are roots of (2) then α, β, γ are divisors of a2 d and as such must be odd integers. Now

                  – b = α + β + γ ⇔ b is odd

and               ac = βγ + γα + αβ ⇔ c is odd

 ⇒                bc is odd. A contradiction.



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