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If $f(x)=ax^2+bx+c$ has an irrational root, if $u=\frac{p}{q}$ be any rational number. a,b,c,p and q are integers. Prove that $\frac{1}{q^2}\le|f(u)|$

My approach $b^2-4ac>0$,$b^2-4ac\ne k^2$, $k$ is a rational number.

$f(u)=a(\frac{p}{q})^2+b(\frac{p}{q})+c$, could not approach after this step.

Samar Imam Zaidi
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    Hint: first show that the assumptions imply that $f(x)$ has no rational roots. In particular, $f(u)\neq 0$. Then prove that $f(u)$ is rational and that it can be written as a fraction with denominator $q^2$. – lulu Oct 23 '18 at 17:59
  • will this be of help? https://proofwiki.org/wiki/Liouville%27s_Theorem_(Number_Theory) – maveric Oct 23 '18 at 18:29

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