How to find square root of 274?

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Answer 1 · 2015-08-18T23:36:48

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

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

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:

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