Question

How to find square root of 274?

Answer 1

Algebraically `sqrt(274)` cannot be simplified much since `274` has no square factors.

You can calculate an approximation `sqrt(274) ~~ 16.552945357`

Explanation:

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

For example:

`sqrt(1008) = sqrt(2*2*2*2*3*3*7) = 2*2*3*sqrt(7) = 12sqrt(7)`

In the case of `274` we have a prime factorisation `274 = 2 * 137`

So we can get `sqrt(274) = sqrt(2*137) = sqrt(2)*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 = 33/2`

Then we can iterate using a Newton Raphson method:

`a_(i+1) = (a_i^2 + n) / (2a_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) = 2p_i q_i`

`p_1 = 33^2 + 274 * 2^2 = 1089 + 1096 = 2185`
`q_1 = 2xx33xx2 = 132`

`p_2 = 2185^2 + 274 * 132^2 = 4774225 + 4774176 = 9548401`
`q_2 = 2xx2185xx132 = 576840`

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

`sqrt(274) ~~ p_2/q_2 = 9548401 / 576840 ~~ 16.552945357`