Let $a,b,c$ be integers, and suppose the equation $$f(x)=ax^2+bx+c=0$$ has an irrational root $r$. Let $u=p/q$ be any rational number such that $|u-r|<1$. Prove that $$\left(\frac{1}{q^2}\le \left|f\left(u\right)\right|\le K\left|u-r\right|\right)$$ for some constant $K$. Deduce that there is a constant $M$ such that $$\left| r\frac{p}{q}\right|\ge\frac{M}{q^2}$$
My work: Let the other root of $f(x)$ be $s$
$|f(u)|=|u-r||u-s|$
For the upper limit of $f(u)$,
$r-1<u<1+r$, value of $u$ is dependent on the value of $r$. We assume that that the value of $r$ and $s$ is finite so value of $|u-s|$ must also be finite.
There must exist some $K$ such that $|u-s|<K$. Multiplying $|u-r|$ we get the desired inequality.
Another way might be:
Let $f(u)=(u-r)(u-s)=k$ (we know $k$ is finite because $f(x)$ is defined for every $x\in \mathbb{R}$, $u-s=\frac{k}{u-r}$ which should be finite. But for $u=r$ $u-r=k=0$, $u-s$ can take any real value which takes us back to square one.
Now, $r+s=-\frac{b}{a}$ which is an integer. $s=-\frac{b}{a}-r$ (By vieta's)
Applying it to the inequality, $u-s=u-\frac{b}{a}-r$ and we know $1<u-r<1$. $$-1-\frac{b}{a}<u-s<1-\frac{b}{a}$$ which is finite so there exists a constant $K$ such that $|u-s|<K$
For the lower limit of $f(u)$, we need to prove $$\frac{1}{q^2}\le \left|\frac{(p-qr)(p-qs)}{q^2} \right|$$ $$\Rightarrow 1 \le|(p-qr)(p-qs)|$$
Will someone proof-read my proof and provide me hints to move forward?