# How do you find the square root of 1414?

Jun 26, 2016

$\sqrt{1414}$ cannot be simplified, but you can calculate approximations.

For example:

$\sqrt{1414} \approx \frac{35347}{940} \approx 37.60319$

#### Explanation:

Factorising $1414$ into prime factors, we find:

$1414 = 2 \times 7 \times 101$

which has no square factors. So its square root has no simpler form.

You can find rational approximations for the square root using a kind of Newton Raphson method.

Given an initial approximation ${a}_{0}$ for $\sqrt{n}$, iterate using the formula:

${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$

In order to make the arithmetic less messy, I prefer to split ${a}_{i} = {p}_{i} / {q}_{i}$ and iterate using the formulae:

${p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2}$

${q}_{i + 1} = 2 {p}_{i} {q}_{i}$

If the resulting ${p}_{i + 1}$ and ${q}_{i + 1}$ has a common factor, then divide both by that factor before the next iteration.

In our example $n = 1414$.

Note that ${37}^{2} = 1369$ and ${38}^{2} = 1444$

So linearly interpolating, choose:

${a}_{0} = 37 + \frac{1414 - 1369}{1444 - 1369} = 37.6 = \frac{188}{5}$

So ${p}_{0} = 188$, ${q}_{0} = 5$

Then:

${p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {188}^{2} + 1414 \cdot {5}^{2} = 35344 + 35350 = 70694$

${q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 188 \cdot 5 = 1880$

These are both divisible by $2$, so do that to get:

${p}_{1 a} = \frac{70694}{2} = 35347$

${q}_{1 a} = \frac{1880}{2} = 940$

If we stopped at this stage we would have:

$\sqrt{1414} \approx \frac{35347}{940} \approx 37.603191489$

Let's try another iteration:

${p}_{2} = {p}_{1 a}^{2} + n {q}_{1 a}^{2} = {35347}^{2} + 1414 \cdot {940}^{2} = 1249410409 + 1249410400 = 2498820809$

${q}_{2} = 2 {p}_{1 a} {q}_{1 a} = 2 \cdot 35347 \cdot 940 = 66452360$

So:

$\sqrt{1414} \approx \frac{2498820809}{66452360} \approx 37.60319135392633158551$

Actually:

$\sqrt{1414} \approx 37.60319135392633134161$