Prove that the values of $x$ for which $x = \frac{x^2+1}{198}$ lie between $\frac{1}{198}$ and $199.99 \overline{49}$. 
Without using a calculator, prove that the values of $x$ for which $x = \dfrac{x^2+1}{198}$ lie between $\dfrac{1}{198}$ and $199.99 \overline{49}$. Then use this result to prove that $\sqrt{2} < 1.41 \overline{421356}$. Finally, is it true that that $\sqrt{2} < 1.41421356$?

For the first part we have that the roots are $99-70\sqrt{2}$ and $99+70\sqrt{2}$ by the quadratic formula. How do I show these roots lie in this range? Also is using the quadratic formula here a good idea?

we have $$\frac{1}{198}<x<199.99\overline{49}=\frac{39599}{198}$$
with $x=\frac{x^2+1}{198}$ we get $1<x^2+1$ which is true.on the other hand we have $$x^2+1<39599$$ this is true since $x^2<198^2<39598$
