# What is the square root of 84?

Apr 25, 2018

$2 \sqrt{21}$

#### Explanation:

$\sqrt{{2}^{2} \left(3\right) \left(7\right)}$

Pull out the pair of twos that are in common and place them on the outside of the radical.

Inside the radical, you are left with $\left(3\right) \left(7\right) = 21$

Hence,

$\sqrt{84} = 2 \sqrt{21}$

Apr 27, 2018

$\sqrt{84} \approx \frac{665335}{72594} \approx 9.1651513899$

#### Explanation:

The square root of $84$ is an irrational number a little larger than $9$.

A square root of a number $n$, is a number $a$ such that ${a}^{2} = n$. Actually every positive number has two square roots, but "the" square root is usually taken to mean the positive one. The two square roots of $84$ are denoted $\sqrt{84}$ and $- \sqrt{84}$.

Note that:

${9}^{2} = 81 < 84 < 100 = {10}^{2}$

Hence:

$9 < \sqrt{84} < 10$

Since $84$ is $\frac{3}{19}$'s of the way between $81$ and $100$, we can approximate $\sqrt{84}$ as $\frac{3}{19} \approx \frac{1}{6}$th of the way between $9$ and $10$, about $9 \frac{1}{6} = \frac{55}{6}$.

In fact we find:

${55}^{2} = 3025 = 3024 + 1 = 84 \cdot {6}^{2} + 1$

Hence $\frac{55}{6}$ is a very efficient approximation to $\sqrt{84}$

Consider the quadratic whose zeros are $55 + 6 \sqrt{84}$ and $55 - 6 \sqrt{84}$:

$\left(x - \left(55 + 6 \sqrt{84}\right)\right) \left(x - \left(55 - 6 \sqrt{84}\right)\right) = {x}^{2} - 110 x + 1$

From this we can define a sequence recursively as follows:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 110 {a}_{n + 1} - {a}_{n}\end{matrix}\right.$

The first few terms of this sequence are:

$0 , 1 , 110 , 12099 , 1330780$

The ratio between successive terms of this sequence tends rapidly towards $55 + 6 \sqrt{84}$

So we find:

$\sqrt{84} \approx \frac{1}{6} \left(\frac{1330780}{12099} - 55\right) = \frac{665335}{6 \cdot 12099} = \frac{665335}{72594} \approx 9.1651513899$