Approximating non-rational roots by a rational roots for a quadratic equation Let $a,b,c$ be integers and suppose the equation $f(x)=ax^2+bx+c=0$ has an irrational root $r$. Let $u=\frac p q$ be any rational number such that $|u-r|<1$. Prove that $\frac 1 {q^2} \leq |f(u)| \leq K|u-r|$ for some constant $K$. Deduce that there is a constant $M$ such that $|r \frac p q| \geq \frac M {q^2}$. (This is useful in approximating the nonrational zeros of a polynomial.)
(I wasn't sure what a non-rational root is. And used intermediate value theorem for $x=u-1$ and $x=u$ but couldn't move further.)
 A: A non-rational root is simply a root that is not a rational number (i.e. it is irrational).
The problem has nothing to do with the intermediate value theorem, it is an exercise in pure algebra. First, note that if $r$ is an irrational root, then the other root $r'$ must also be irrational (assume that $r'$ is rational; since $r+r'= - \frac b a$ which is rational, this would imply $r$ rational which is a contradiction). We shall use this in a moment.
Now $|f(u)| = \frac {|a p^2 + bpq + c q^2|} {q^2}$, so let us show that the numerator is $\geq 1$. This numerator is an integer, so it is either $0$, or $\geq 1$. If it is $0$, then divide by $q^2$ and get $f(\frac p q)=0$. But this means that $f$ has a rational root, impossible according to the first paragraph. So, $\frac 1 {q^2} \leq |f(u)|$.
Also, since $f(r)=0$, $|f(u)| = |f(u) - f(r)| = |a(u^2-r^2) + b(u-r)| = |u-r| \cdot |a(u+r)+b|$. Now, it is not clear what is meant by "constant" in the statement of the problem: constant with respect to what? I am going to assume that one means "constant with respect to u" (considering $r$ fixed). Then $|u+r| = |-r -u| = |-2r +r-u| \leq 2|r| + |r-u| \leq 2|r| + 1$ and then $|a(u+r) + b| \leq |a| \cdot |u+r| + |b| \leq |a|(2|r|+1)+|b|$. If you call this last number $K$, then $|f(u)| \leq K |u-r|$.
Finally, take the endpoints of this inequality, $\frac 1 {q^2} \leq K|u-r|$, multiply them by $|u|$ and get $\frac {|u|} {q^2} \leq K |u^2 -ur| \leq Ku^2 + K|ur|$. Therefore $|\frac p q r| \geq \frac {|u|} {K q^2} - u^2$ and, if you let $M=\frac {|u|} K - u^2 q^2$, this can be rewritten as $|\frac p q r| \geq \frac M {q^2}$.
