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Given the formula for the 4th degree polynomial, is it true that a root is a natural only when all the radicals inside the formula are rational numbers?

Edit1: The coeficients are whole numbers.

https://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg

Robert Puscasu
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    Strange things can happen with radicals. $6+4\sqrt2$ and $6-4\sqrt2$ are both irrational, but $\sqrt{6+4\sqrt2}+\sqrt{6-4\sqrt2}=4$. – Gerry Myerson May 02 '22 at 11:14
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    @GerryMyerson You're right. Your answer helped me solve my problem. Thanks. – Robert Puscasu May 03 '22 at 09:06
  • Let me see if get this right. If i just write the radicals who generate imaginary numbers in exponential form using Euler's formula then they will easily cancel each other? https://en.wikipedia.org/wiki/Euler%27s_formula? – Robert Puscasu Jul 07 '22 at 03:02
  • I'm not sure I understand your question. We were talking about natural numbers and irrationals, not about real numbers and complex ones. The numbers $\sqrt{6\pm4\sqrt2}$ are both real, so I don't see any room for applying Euler. – Gerry Myerson Jul 07 '22 at 03:20

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