# How to find square root of 274?

##### 1 Answer
Aug 18, 2015

Algebraically $\sqrt{274}$ cannot be simplified much since $274$ has no square factors.

You can calculate an approximation $\sqrt{274} \approx 16.552945357$

#### Explanation:

If the radicand of a square root has square factors, then it can be simplified.

For example:

$\sqrt{1008} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7} = 2 \cdot 2 \cdot 3 \cdot \sqrt{7} = 12 \sqrt{7}$

In the case of $274$ we have a prime factorisation $274 = 2 \cdot 137$

So we can get $\sqrt{274} = \sqrt{2 \cdot 137} = \sqrt{2} \cdot \sqrt{137}$

but that's not really simpler.

If we want to approximate the value of $\sqrt{274}$ then that's a rather different question.

Note that ${16}^{2} = 256$ and ${17}^{2} = 289$, so let our first approximation be halfway between:

${a}_{0} = \frac{33}{2}$

Then we can iterate using a Newton Raphson method:

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

where $n = 274$.

Actually, I prefer to keep the numerator and denominator of the rational approximation as separate integers and iterate as follows:

$n = 274$
${p}_{0} = 33$
${q}_{0} = 2$

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

${p}_{1} = {33}^{2} + 274 \cdot {2}^{2} = 1089 + 1096 = 2185$
${q}_{1} = 2 \times 33 \times 2 = 132$

${p}_{2} = {2185}^{2} + 274 \cdot {132}^{2} = 4774225 + 4774176 = 9548401$
${q}_{2} = 2 \times 2185 \times 132 = 576840$

Stop when you think you have enough significant digits, then divide to get your approximation:

$\sqrt{274} \approx {p}_{2} / {q}_{2} = \frac{9548401}{576840} \approx 16.552945357$