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Here, discriminant `D =b^(2) - 4ac.` Suppose the roots are rational. Thus , D will be a perfect square. <br> Let `b^(2) - 4 ac = d^(2)` . Since a, b and c are odd integers, d will be odd .Now, <br> `b^(2) - d^(2) = 4ac` <br> Let b = 2k + 1 and d = 2m + 1 . Then <br> `b^(2) - d^(2) = (b -d) (b + d)` <br> `2(k -m) 2(k + m + 1)` <br> Now, either (k - m) or (k + m + 1) is always even. Hence, `b^(2)-d^(2)` is always a multiple of 8. But , 4ac is a multiple of 4 (not of 8 ), Which is a contradiction. Hence, the roots of `ax^(2) + bx + c =0` connot be rational. **What is Real polynomial, complex polynomial and degree of a polynomial?**

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