How do you find the square root of 361?

##361 = 19^2##, so ##sqrt(361) = 19##.

See explanation for a few methods...

Prime Factorisation One of the best ways to attempt to find the square root of a whole number is to factor it into primes and identify pairs of identical factors. This is a bit tedious in the case of ##361## as we shall see:

Let's try each prime in turn:

##2## : No: ##361## is not even. ##3## : No: The sum of the digits is not a multiple of ##3##. ##5## : No: The last digit of ##361## is not ##0## or ##5##. ##7## : No: ##361 -: 7 = 51## with remainder ##4##. ##11## : No: ##361 -: 11 = 32## with remainder ##9##. ##13## : No: ##361 -: 13 = 27## with remainder ##10##. ##17## : No: ##361 -: 17 = 21## with remainder ##4##. ##19## : Yes: ##361 = 19*19##

So ##sqrt(361) = 19##

Approximation by integers ##20*20 = 400##, so that's about ##10##% too large.

Subtract half that percentage from the approximation: ##20 - 5##% ##= 19##

The "half that percentage" bit is a form of Newton Raphson method.

Try ##19*19 = 361## Yes.

Hmmm, I know some square roots already I know ##36 = 6^2## and ##sqrt(10) ~~ 3.162##, so:

##sqrt(361) ~~ sqrt(360) = sqrt(36) * sqrt(10) ~~ 6 * 3.162 ~~ 19##

Try ##19*19 = 361## Yes

Memorise Hey! I know that already: ##361 = 19^2##

Knowing a few squares is useful for all sorts of mental calculation, so I would recommend memorising them a bit. In fact you can multiply two odd or two even numbers using squares, adding, subtracting and halving as follows:

##a xx b = ((a+b)/2)^2 - ((a-b)/2)^2##

For example:

##23 * 27 = 25^2 - 2^2 = 625 - 4 = 621##