How do you find the square root of 529?

3 Answers
Sep 8, 2015

Answer:

Check for divisibility by a perfect square to simplify. You will find that #23^2 = 529#

Explanation:

When we try to simplify a square root, we look for perfect square factors.

Do this by testing perfect squares until you get to a number whose square if greater that #529# (for example #25^2 = 625#)

So we test #2^2 = 4#. Clearly #4# is not a factor of #529# (nor is any other even number.

Test #3^2 = 9# which is not a factor. Skip #4^2# because it is even.

Obviously #5^2=25# is not a factor. Skip #6^2#.
Keep going . . .

#23^2 = 529# so we're finished.

#sqrt529 = 23#

May 17, 2017

Answer:

Use a mixture of methods to find #sqrt(529) = 23#

Explanation:

There are quite a few different ways to find square roots.

Here's a bit of a mish-mash for this particular example...

Given #529#, first split off pairs of digits starting from the right hand side to get:

#5|29#

Next, note that #5# lies between two square numbers #2^2 = 4# and #3^2 = 9#

We can approximate where #sqrt(5)# lies between #2# and #3# by linearly interpolating. What's that? We approximate the part of the graph of #x^2# between #x=2# and #x=3# using a straight line to get our approximation.

Since #5# is #1/5# of the way between #2^2=4# and #3^2=5#, we can approximate #sqrt(5)# by #2+1/5 = 2.2#

Hence a good first approximation for #sqrt(500)# is #22#.

How about #529#?

We can use the Babylonian method to find a better approximation.

Given a positive number #n# and an approximation #a_i# to its square root, a better approximation is given by the formula:

#a_(i+1) = (a_i^2+n)/(2a_i)#

So putting #n=529# and #a_0=22#, we find:

#a_1 = (a_0^2+n)/(2a_0) = (22^2+529)/(2*22) = (484+529)/44 = 1013/44 = 23.02bar(27)#

That looks suspiciously close to #23# so try:

#23^2 = 529#

Thus #sqrt(529) = 23#

Jun 30, 2017

Answer:

#sqrt529 =23#

Explanation:

Do a really rough estimate first using squares of multiples of #10#

#20^2 = 400 and 30^2 = 900#

#529# lies between #400 and 900#

so #sqrt529# will lie between #20 and 30#

Now look at the last digit ...#9#
There are only two numbers whose squares end with a #9#,

#3 and 7 rarr " "3^2 =9 and 7^2 =49#

So the possibilities are #23 and 27#

However, #25^2 = 625 and 529# is less than #625#

My first guess would therefore be #23#

Multiplying confirms that #23 xx 23 = 529#

Hence #sqrt529 =23#