What is the square root of #31# ?

3 Answers
Sep 20, 2017

#5.56#

Explanation:

All you need to do is take a calculator then press this sign #sqrt# then put the number under it.

Sep 20, 2017

#+-5.56776436283#

Explanation:

#because sqrt31 = +-5.56776436283#
However if you are using this in algebra i would leave it as #sqrt31# as it is easier than writing #+-5.56776436283#.

Sep 20, 2017

#sqrt(31)# is an irrational number approximately equal to:

#120041/21560 ~~ 5.5677644#

Explanation:

Since #31# is a prime number, its square root is an irrational number.

Actually, every non-zero number has two square roots.

In the case of #31# they are written #sqrt(31)# and #-sqrt(31)#.

#sqrt(31)# is the called the principal square root and is the positive one.

#-sqrt(31)# is the negative square root.

When people say "the square root", they generally mean "the principal square root", i.e. the positive one.

Note that:

#5^2 = 25 < 31 < 36 = 6^2#

So the (positive) square root of #31# is somewhere between #5# and #6#.

Using a calculator:

#sqrt(31) ~~ 5.56776436283#

but we don't need a calculator to find rational approximations to #sqrt(31)#.

There are at least #25# methods we could use, but one effective one is the Babylonian method:

If #a# is an approximation to #sqrt(n)#, then a better one is:

#(a^2+n)/(2a)#

Since #31# is about half way between #25=5^2# and #36=6^2#, a good first approximation to #sqrt(31)# is #11/2#.

So a better approximation is:

#((11/2)^2+31)/(2*11/2) = (121/4+124/4)/11 = 245/44#

Repeat to get a better approximation:

#((245/44)^2+31)/(2*245/44) = 120041/21560 ~~ 5.56776438#