How do you find the square root of any number?

## How do you find the square root of 80

# 3 Answers

The simplest way is to make a series of educated guesses until you get as close as you need to be. What is the highest number you can think of that, when squared, is still less than 80? What is the lowest number that, when squared, will be higher than 80? 8 squared is 64, and 9 squared is 81. You now know the answer is between 8 and 9. Now for the bad news: it won't be a rational number, and to get closer, it gets messy.

We could try 8.5 squared, which is 72.25. Still low, so now we know it's between 8.5 and 9.

How about 8.8? That squared is 77.44. Still low. Must be between 8.8 and 9.

Eventually you will find two numbers close enough that when they are squared, the result, rounded to the nearest ___ (depends upon how close you need to come), will be 80. For example, 8.94 squared is 79.9236, which rounded to the nearest one is 80, as is 8.95 squared (which is 80.1025).

If you need it closer than that, pick a number between 8.94 and 8.95 and continue from there in the same way.

In russian grade schools, besides learning addition, subtraction, multiplication, and long division algorithms, they also teach paper and pencil square roots: Unfortunately this answer system puts blank lines when you do a carriage return so do your best to understand:

8 . 9 4 4 2 ...

Sqrt 8 0 . 0 0 0 0 0 0 0 0 0 0 ...

6 4

1 6 0 0

1 5 2 1 = (20*8+9)*9 so 9, next digit is 9

7 9 0 0

7 1 3 6 = (20*89+4)*4, next digit is 4

7 6 4 0 0

7 1 5 3 6 = (20*894+4)*4, next digit is 4

4 8 6 4 0 0

3 5 7 7 6 4 = (20*8944+2)*2, next digit is 2

1 2 8 6 3 6 0 0

...

Do a prime factorization of the number that you want to take the square root of. Using Margaret's example of 80:

80 = 40 * 2 (2 is prime, so you're done with that number - I've underlined it)

Now factor the 40; 40 = 20 * 2

Factor the 20; 20 = 10 * 2

Factor the 10; 10 = 5 * 2

Now you can write 80 as a product of prime factors: 80 = 2 * 2 * 2 * 2 * 5

Because you're taking the square root of 80, you want to look for numbers in your prime factorization that occur twice. In this example you have 2 pairs of 2's. I've highlighted one pair and italicized the other pair so you can visualize it: 80 =
**2 * 2 *** *2 * 2 ** 5.

If a factor occurs twice, you can pull it out of the square root; 80 = 2 * 2 * sqrt (5) = 4 sqrt (5), or 4 times the square root of 5.

Another example: square root of 45.

45 = 9 * 5 (5 is prime, but 9 is not so you have to factor 9 further - when you factor 9 further, you get that 9 = 3 * 3)

45 = 3 * 3 * 5

Pull the pair of 3's out from under the square root:

square root (45) = 3 * square root (5)

If you wanted to take the cube root of a number, you would look for factors that occur three times. For our example of 80:

80 = 2 * 2 * 2 * 2 * 5 ; 2 occurs three times. You can pull those out of the cube root:

80 = 2 * cube root (2 * 5) = 2 * cube root (10)

## Comments

Brilliantly simple. Whether or not you can leave it as something like:

4 * sq rt (5)

depends upon why you want to find the square root in the first place. But it generally helps if you can simplify first. Thanks for pointing this out.

Comment