How to find square root of 274?

1 Answer
Aug 18, 2015

Answer:

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#