# How do you simplify the square root of 95?

Mar 17, 2018

$\sqrt{95} \approx \frac{18495361}{1897584} \approx 9.746794344809$

#### Explanation:

The prime factorisation of $95$ is:

$95 = 5 \cdot 19$

Since this contains no square factors, $\sqrt{95}$ is already in simplest form. There are no factors that can be moved outside the radical.

As a continued fraction, we find:

sqrt(95) = [9;bar(1,2,1,18)] = 9+1/(1+1/(2+1/(1+1/(18+1/(1+1/(2+1/(1+1/(18+...))))))))

Hence an efficient rational approximation for $\sqrt{95}$ is:

[9;1,2,1] = 9+1/(1+1/(2+1/1)) = 39/4

For a fun way to find better approximations to $\sqrt{95}$, consider the quadratic with zeros $39 + 4 \sqrt{95}$ and $39 - 4 \sqrt{95}$:

$\left(x - 39 - 4 \sqrt{39}\right) \left(x - 39 + 4 \sqrt{95}\right) = {\left(x - 39\right)}^{2} - 1520$

$\textcolor{w h i t e}{\left(x - 39 - 4 \sqrt{39}\right) \left(x - 39 + 4 \sqrt{95}\right)} = {x}^{2} - 78 x + 1$

Based on this, define a sequence recursively by:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 78 {a}_{n + 1} - {a}_{n}\end{matrix}\right.$

The first few terms of this sequence are:

$0 , 1 , 78 , 6083 , 474396 , 36996805$

The ratio between successive terms will converge rapidly to $39 + 4 \sqrt{95}$.

Hence we find:

$\sqrt{95} \approx \frac{1}{4} \left(\frac{36996805}{474396} - 39\right) = \frac{1}{4} \left(\frac{18495361}{474396}\right) = \frac{18495361}{1897584}$

$\textcolor{w h i t e}{\sqrt{95}} \approx 9.746794344809$