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I know about algebraic numbers and transcendental numbers. How the roots of a polynomial with irrational coefficients are classified. Are they transcendental?

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They are not necessarily transcendental: Take any polynomial with rational coefficients and multiply its coefficients by $\sqrt{2}$. This is a polynomial with irrational coefficients, which has the same roots as the original. –  J. J. Feb 24 '11 at 15:20
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up vote 4 down vote accepted

The roots of a polynomial with algebraic coefficients are all algebraic, and a monic polynomial whose roots are all algebraic has algebraic coefficients.

So a monic polynomial with some transcendental coefficient must have at least one transcendental root (and vice versa), but it can also have algebraic roots (for example, $0$ is a non transcendental root of $X^2- \pi X = 0$).

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If you change $\sqrt{2}$ to $\pi$ in J.J.'s comment you need not have a transcendental root even if the coefficients are transcendental. –  Ross Millikan Feb 24 '11 at 15:31
    
right, I forgot to keep the monic condition. –  mercio Feb 24 '11 at 15:33
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