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?
$\endgroup$1 Answer
$\begingroup$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 $x^2<198^2<39598$
$\endgroup$ 4
