How to use $t(29/\sqrt{2})<0$ where $t(x)=x^2-41x+420$ to prove that $41/29<\sqrt{2}<42/29$??

Question

So I was investigating different ways to approximate $\sqrt{2}$.

Here's my latest:

$$Let:t(x)=x^2-41x+420$$

then the roots of $t(x)$ are $20$ and $21$.

I showed that then $t(x)=(x-20)(x-21)$ and know I have to show that $t\left(\frac{29}{\sqrt{2}}\right) \lt 0$. I used the fact that I got from another question that $\frac{40}{29} \lt \sqrt{2}< \frac{42}{29}$ and using a table I proved that $t\left(\frac{29}{\sqrt{2}}\right) \lt 0$.

Now I need to conclude from $t\left(\frac{29}{\sqrt{2}}\right) \lt 0$ we have $\frac{41}{29} \lt \sqrt{2}< \frac{42}{29}$

But I don't know how to proceed!

Thank you so much!

Answer 1

Squaring is the best method to prove the inequality. However, there is a detour, which involves squaring and nastier computation (and somewhat magical factorization), that may be what you're looking for.

Substitute $x = \frac{29}{\sqrt 2}$ in $t(x) = x^2 - 41x + 420$ to get \begin{align} t\left(\frac{29}{\sqrt 2}\right) = \frac{29^2}{2} - 41 \cdot \frac{29}{\sqrt 2} + 420 & < 0 \\ 29^2 + 42 \cdot 20 & < 41 \cdot 29 \cdot \sqrt 2 \\ \frac{41^2}{41 \cdot 29} & < \sqrt 2 \\ \frac{41}{29} & < \sqrt 2. \end{align}

Answer 2

If $(x-20)(x-21)\lt 0$ then you must have $20\lt x \lt 21$ and you can't prove anything stronger from this inequality.

So if $t\left(\frac {29}{\sqrt 2}\right)\lt 0$ you have $$20\lt\frac {29}{\sqrt 2}\lt 21$$ If you multiply this through by $\frac 2{29}$ you get: $$\frac{40}{29}\lt \sqrt 2\lt \frac {42}{29}$$

You can't do better than this without a different method.