Suppose we are interested in producing a prime near a very large number $M$. It is convenient to not bother testing for primality the multiples of small primes, like the even numbers! For the remaining numbers, test, one after the other, candidates using a "probable prime" test such as Miller-Rabin.
Miller-Rabin is very fast, and can be used to quickly rule out candidates.
If we are not too fussy, we can accept such a probable prime as prime. There is a possibility of being wrong, but if we make the probability of error less than say $10^{-100}$, effectively we have certainty.
Or else, after locating a number which is "almost certainly prime" via Miller-Rabin, one can use one of the relatively expensive primality tests to make sure.
First, it's important to note that this only works for roots of perfect squares.
Start by examining what happens to multiples of 10 when you square them:
$$10^{2}=100$$
$$20^{2}=400$$
$$30^{2}=900$$
$$40^{2}=1600$$
$$50^{2}=2500$$
$$60^{2}=3600$$
$$70^{2}=4900$$
$$80^{2}=6400$$
$$90^{2}=8100$$
So, because they're multiples of 10, the squares all wind up being multiples of 100. Yes, adding the numbers 1 through 9 will increase this, but even, say, $79^{2}$ can't be greater than $80^{2}$. Therefore, if a perfect square is between 4900 and 6400, then it's root must be 70-something.
That explains getting the tens digit, but what about the units digit?
Look at the patterns of the squares of the one-digit numbers, paying particular attention to the units digit:
$$1^{2}=\textbf{1}$$
$$2^{2}=\textbf{4}$$
$$3^{2}=\textbf{9}$$
$$4^{2}=1\textbf{6}$$
$$5^{2}=2\textbf{5}$$
$$6^{2}=3\textbf{6}$$
$$7^{2}=4\textbf{9}$$
$$8^{2}=6\textbf{4}$$
$$9^{2}=8\textbf{1}$$
Notice the units digits of the squares go up 1, 4, 9, 6, 5, and then go back down in reverse order, 6, 9, 4, 1.
So, a perfect square ending in 1, must have a root ending in 1 or 9. A perfect square ending in 4 must have a root ending in 2 or 8, and so on.
For the perfect square of a 2-digit number, you've already narrowed it to 2 possibilities.
The puzzling part about this is probably how you choose between the 2. The standard method is to learn a quick trick for squaring numbers ending in 5.
To square a 2-digit number ending in 5, you take the tens digit, multiply it by 1 greater than itself, and put 25 to the left of that result. For 25, then tens digit is 2, so we multiply 2 by 3 to get 6, stick a 25 on the end, and we have 625. To square 35, we'd multiply 3 (the tens digit) by 4 (one more than the tens digit) to get 12, add the 25 to the end, giving us 1225.
For use in determining roots, wait until you get to the part where you have 2 choices. For example, given 784, you can work out that the tens digit of the answer is in the 20s, and that the units digit must be 2 or 8. How do you decide?
Well, using the trick you just learned, you can quickly calculate that 25 squared is 625. Is 784 more or less than 625? Obviously, it's more, which means the square root must be the greater of the 2 choices, so the square root of 784 must be 28.
The article to which you link tries a bit of a short cut by not teaching you the full square-a-number-ending-in-5 trick, but in my opinion, it just seems to make the whole trick more confusing.
Why does the square-a-number-ending-in-5 trick work?
Looking at this algebraically, you can think of squaring a number ending in 5 as:
$$(10x+5)^2=(10x+5)(10x+5)$$
This multiplies out to:
$$100x^{2}+100x+25$$
We can factor this as:
$$100(x^2+x)+25$$
And we can even factor it once more:
$$100\{x(x+1)\}+25$$
There's the trick itself laid bare. Multiply x by 1 greater than itself, multiply it by 100 to move the answer over 2 places to the left, and then add 25.
Best Answer
Many mental calculations are easier the more facts you memorize. Knowing the squares up to $31^2=961$ can make it easier, reducing the need to go all the way to single digits. Also the fact that $(10n+5)^2=100n(n+1)+25$, for example $65^2=(6\cdot7)25=4225$. If you break it all the way to individual digits, I find it easier to keep track of my place if I start with the most significant digits and add as I go, so $359^2=300^2+2\cdot300\cdot50+\ldots =90000+30000+\ldots$ $=120000+ 50^2+\ldots = 122500+2\cdot300\cdot9+\ldots=127900+2\cdot50\cdot9+9^2$ where for many purposes you can quit early when you have the precision needed.