Finding the square root of a big number, like 676?

Question

I am having trouble understanding and finding the square roots of large numbers. How would I go about finding this number efficiently?

Answer 1

Square root of $676$ is the number such that multiplied by itself will give you 676. This number is denoted as $\sqrt{676}$.

Note that generally speaking, for a non-zero real number $p$ there are two numbers which squared will give you $p$. They differ by the sign. In order to ensure uniqueness of the square root, the general convention is for positive numbers to denote square another positive number as its square root.

If you want to calculate a square root of a large number, it is often helpful to represent this large number as a product of smaller numbers, e.g.

$$ 676 = 4\cdot 169 = (2\cdot 2) \cdot (13\cdot 13) = 2^2\cdot 13^2 $$

Therefore

$$ \sqrt{676} = \sqrt{4\cdot 169 } = \sqrt{2^2}\cdot \sqrt{ 13^2} = 26 $$

---

EDIT: Thanks to the comments, I am able to get rid of unclarity in my original post.

Answer 2

I'm going to take a different tack than the earlier answers.

First, most numbers don't have whole number square roots, so I'm going to talk about getting close to a square root - without finding its factors, and even without a calculator, if you can manage a little mental arithmetic (or manual arithmetic). Secondly, if it is a problem that has a whole number square root, this gives you a way to quickly guess at it.

Step 1: mentally ignore an even number of digits to the right, so you only look at 1 or 2 digits to the left:

$\qquad 6\text{ _ _}$

Step 2: Find the largest integer whose square is smaller than this new number (in this case, $2^2=4$ and $3^2=9$, so it's $2$.

This is the first digit of the square root.

Step 3: It will have as many additional digits as half the number of digits you "left off" above (which was two: "$\text{ _ _}$"), so in this case it has one more digit.

Step 4: Now if the original is the square of a whole-number, you can see the second digit must be either $4$ or $6$, since those are the only end-digits that produce a $6$ in their square (the squares of 0 to 9 end in 0,1,4,9,6,5,6,9,4,1). So the square root in that case must be either 24 or 26. If you then know the trick for squaring numbers ending in 5 ($25^2 = 20\times 30+25=625$), it's immediately clear it must be $26$, which is easy to find the factors of, so this also speeds up the identification of the factors of the original square number.

But as I was suggesting, most numbers you encounter don't have whole-number square roots. So one way to get close is to guess at it and then improve the guess by averaging the guess with the original number divided by the guess.

Since we know the first digit is 2, the number is at least 20 and below 30. So let's start by guessing "25".

Then we improve the guess by averaging $25$ and $\frac{676}{25}=27\frac{1}{25}$. The average is $26\frac{1}{50}$ (halfway between 25 and 27 plus half the leftover fraction). However, these second-guess averages are always a little too big for the square root, so while $26\frac{1}{50}=26.02$ is a good next guess, it might help to round down to $26$ ... whereupon we find when we try to calculate the next guess that we have hit it exactly. In cases where it's not a whole number we can hit exactly, a couple of rounds of such dividing and averaging can get really close quite rapidly.

Let's try this on a bigger problem which we're told has a whole-number square root: $375769$

1. Leave off an even number of digits on the right, to leave one or two digits:

   $3 7 _ _ _ _$

2. The first digit must be $6$ ($6^2=36$). We left off 4 digits, so the square root has two more digits.

3. Since $6^2$ is quite close to $37$, guess "600".

4. we want to average 600 and $\frac{375769}{600}$ ... but we don't have to get it exactly, so in that second term let's round off the numerator to the nearest 100: $\frac{375800}{600}=\frac{3758}{6}=626.\text{<something>}$. Now average with the 600 we started with to get our second guess: $613.\text{<and a bit>}$ ... but this number will be slightly too high.

5. Clearly the last digit must be either 3 or 7 (since they are the only digits whose square ends in 9).

6. so we need a number a bit smaller than $613.\text{<and a bit>}$ that either ends with 3 or 7. The obvious one to try is $613$ ... which is the required number. (If 613 wasn't the square root, it would either be 607 or we made an error somewhere. With practice this sort of thing can be very rapid.)

---

Note also that there are only 22 three digit whole-number squares. It's surprisingly quick to learn them all (there are several patterns that can be exploited to speed that up); this will speed up identifying square roots and can provide numerous shortcuts to calculations like the one we just did -- and to other calculations as well (e.g. $17\times 19=18^2-1=323$).

Answer 3

Hint: Find the prime factorization, then divide all exponents by 2. This is equivalent to doing $$\sqrt n=\sqrt{p_1^{a_1}p_2^{a_2}\cdots}={p_1^{a_1/2}p_2^{a_2/2}\cdots}$$