If $p$ and $q$ are odd integers, then the equation $x^{2}+2px+2q=0$
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has no integral root
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has no rational root
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has no irrational
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has no imaginary root.
Discriminant D = (2p)^2 - 4(1)(2q) = 4p^2 - 8q = 4(p^2 - 2q). For roots to be rational, D must be a perfect square. Since p is odd, p^2 is odd. 2q is even. p^2 - 2q is odd. 4 * (odd) is never a perfect square (it is 4 times an odd number). Thus, roots cannot be rational.
If the equation x squared plus 2px plus 2q = 0 has an integral root m, then m squared must be even since it equals 2 times the quantity p minus m, meaning m itself is even and p minus m is odd. Because m is even, m squared is a multiple of 4, which makes the even number 2 times the quantity p minus m a multiple of 4, forcing the odd number p minus m to be even, which is a contradiction. Since a monic polynomial with integer coefficients cannot have rational roots that are not integers, the equation has no rational root.