# What is the square root of 90?

Mar 2, 2018

$\sqrt{90} = 3 \sqrt{10} \approx \frac{1039681}{109592} \approx 9.48683298051$

#### Explanation:

$\sqrt{90} = \sqrt{{3}^{2} \cdot 10} = 3 \sqrt{10}$ is an irrational number somewhere between $\sqrt{81} = 9$ and $\sqrt{100} = 10$.

In fact, since $90 = 9 \cdot 10$ is of the form $n \left(n + 1\right)$ it has a regular continued fraction expansion of the form [n;bar(2,2n)]:

sqrt(90) = [9;bar(2,18)] = 9+1/(2+1/(18+1/(2+1/(18+1/(2+1/(18+...))))))

One fun way to find rational approximations is using an integer sequence defined by a linear recurrence.

Consider the quadratic equation with zeros $19 + 2 \sqrt{90}$ and $19 - 2 \sqrt{90}$:

$0 = \left(x - 19 - 2 \sqrt{90}\right) \left(x - 19 + 2 \sqrt{90}\right)$

$\textcolor{w h i t e}{0} = {\left(x - 19\right)}^{2} - {\left(2 \sqrt{90}\right)}^{2}$

$\textcolor{w h i t e}{0} = {x}^{2} - 38 x + 361 - 360$

$\textcolor{w h i t e}{0} = {x}^{2} - 38 x + 1$

So:

${x}^{2} = 38 x - 1$

Use this to derive a sequence:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 38 {a}_{n + 1} - {a}_{n}\end{matrix}\right.$

The first few terms of this sequence are:

$0 , 1 , 38 , 1443 , 54796 , 2080805 , \ldots$

The ratio between successive terms will tend to $19 + 2 \sqrt{90}$

Hence:

$\sqrt{90} \approx \frac{1}{2} \left(\frac{2080805}{54796} - 19\right) = \frac{1}{2} \left(\frac{1039681}{54796}\right) = \frac{1039681}{109592} \approx 9.48683298051$