How do you find the square root of 52?

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Answer 1 · 2015-09-12T20:52:18

#sqrt(52) = 2sqrt(13) ~~ 7.21110#

If #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)#, so:

#sqrt(52) = sqrt(2^2*13) = sqrt(2^2)sqrt(13) = 2sqrt(13)#

If you want to calculate an approximation by hand, you could follow the advice I gave for #sqrt(28)# in http://socratic.org/questions/how-do-you-find-the-square-root-28

Using the method described there:

Let #n = 52#, #p_0 = 7#, #q_0 = 1#

Then:

#p_1 = 7^2+52*1^2 = 49+52 = 101#
#q_1 = 2*7*1 = 14#

#p_2 = 101^2+52*14^2 = 10201+10192 = 20393#
#q_2 = 2*101*14 = 2828#

Stopping at this point, we get a result accurate to #5# decimal places:

#sqrt(52) ~~ 20393/2828 ~~ 7.21110#