How do you calculate the square root of a number, without using a calculator.?

Trial and error:

1) 41 is between 36 (6 squared) and 49 (7 squared) so its square root is between 6 and 7.

2) try 6.5 x 6.5 = 42.25. Root of 41 must be just under 6.5

3) try 6.4 x 6.4 = 40.96. Root of 41 is just a tiny bit above 6.4.

4) try 6.41 x 6.41 = 41.0881. Root of 41 is a very tiny bit below 6.41.

5) try 6.405 x 6.405

etc

The method seems long, but is easy when you get used to it.

Example square root of 6,558,721

Divide the number up in bits of two starting from the right so we then have

6,55,87,21

Take the first bit on the left, in this case 6 and find the highest square of a number that will go into 6. Even if this was a double number it could not be higher than 99 so the highest square you have to consider is 9 squared = 81

answer 2...........2 squared is 4

Show as follows, take the 4 from the 6 = 2, bring down the next pair and also write the 2 on the top.

.......2

2.....6, 55, 87, 21

......4

.......255

Double the 2, gives us 4. But think of it as 40 something.

......2

2.....6, 55, 87, 21

......4

4*....255

Now find the largest number that can replace the * with the full number being multiplied by * with the answer being less than 255

43 times 3 = 129

44 times 4 = 176

45 times 5 = 225

46 times 6 = 276

So it is 5. Make the 4* into 45 and also put the 5 up on the top

.......2 5

2.....6, 55, 87, 21

......4

45....255

.......225

......... 3087

As above, deduct 225 from 255 and then bring down the next pair of numbers

Now we have 25 on the top line. Double it makes 50. Add a * as before to give 50*

So now we have a number of 500 and something up to 509.

Do the same as before 501 times 1, 502 time 2 etc

Looks like maybe 506 times 6

That equals 3036. Looks good. 507 times 7 would be 3549, too big.

We now have

.......2 5 6 With that new 6 added to the top line

2.....6, 55, 87, 21

.....4

45....255

.......225

506....3087

........ 3036

We now do the sum of 3087 minus 3036 and bring down the next pair.

......2 5 6

2.....6, 55, 87, 21

.......4

45...255

......225

506....3087

.........3036

.............5121

On the top line we now have 256. Double it and use it as before with the * at the end

.......2 5 6

2......6, 55, 87, 21

........4

45.....255

.........225

506......3087

...........3036

512*.........5121

This one looks easy 5021 times 1 = 5021. So put in the 1. We now have

.......2 5 6 1 With that 1 added to the top line

2..... 6, 55, 87, 21

........4

45 .....255

.........225

506.......3087

.............3036

5121.........5121

.................5121

0

At the top we have 2561. Try your calculator. We have the correct answer.

I used as an example a number that would give a whole number answer. If the answer was not a whole number, you could just keep going after the decimal point as far as you want.

The last time I had to do this must have been about 50 years ago. It is surprising what one can remember.

Please excuse all the ..........

Seems Yahoo does not like leading spaces. Plus I tried to put underlines in the relevant places and that did not work.

I forgot but I think this is how it goes.

one example would be sq root of 8.

You start off by finding a number that you can divid the number 8 by. This number has to give you one number that can be perfect square and another number that you can't. Though if you get two perfect square I think, you have to do it again.

so 8 would be 2 and 4. 2 can't be perfect squared while 4 can.

so sq root of 8 is 2sqroot of 2 which equals sq root of 8.

check it on a calculator.

now for 41 well thats a hard one since there are no numbers that you divide into 41 can be perfect squared. Other than that, yea trail and error works best for 41.

I hope this helps.

Yes you can calculate the Square Root of anumber with out using a calculator.

The ancient Babylonians had a nice method of computing square roots

that can be applied using only simple arithmetic operations. To find

a rational approximation for the square root of an integer N, let k

be any number such that k^2 is less than N. Then k is slightly less

than the square root of N, and so N/k is slightly greater than the

square root of N. It follows that the average of these two numbers

gives an even closer estermate.

. . . . . . . . . k+N/k

new_k = -----------

. . . . . . . . . 2

See link to Acient Square Roots

Obviously mathmaticians way back when had to figure out a way to do it but I've asked several of my math teachers the same thing and they said there is no way to do it. Once again, there HAS to be a way to do it but I'm sure that it involves a complicated formula or a good deal of division that no normal person in there right mind would care to do. It's really not needed anyways since the majority of teachers will let you use a calculator on tests that require the square root of a number to be found

I was at school when calculators just came out, so we were brought up on slide rules and 4-figure tables.

In those days we used logarithms to work out squares and roots.

To work out the square root of 41, look up the logarithm (base 10 will be OK), this gives 1.612. i.e 10 to the power 1.612 is 41.

To find the square root just halve the index, so half of 1.612 is 0.806.

So the square root of 41 is 10 to the power 0.806.

Back to those 4 figure log tables to look up the inverse log and the answer is 6.397. Compared to the calculator answer of 6.403 it is 0.006 out, but thats the modern luxury of 8 decimal places on a calculator.

There is also a way of working it out on a slide rule, but mine went years ago and I can't quite remember how to do it.

It's done best by way of diagrams using circles and triangles (which are half a square) that are drawn to scale and can be measured accurately.

But if you want to get complicated, you can throw in Pythagoras' theorem for the length of a hypoteneuse when the other two sides are of equal length - but don't forget to draw the cricle which has that hypoteneuse as a tangent and the point of right-angle as the centre of the circle.

Then it's easy....and you can work out areas, value of pi, logs, exponential functions....wooohoooooo.

There is a way to extract a square root similar to doing a long devision. I can't remember how to do it now. But here is the link for it.

http://www.qnet.fi/abehr/Achim/Calculators_SquareR...

Ask someone who will use a calculator?

K

go to this page http://www.homeschoolmath.net/teaching/square-root...