how to find square root

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How to calculate a square root without a calculator
and should your child learn how to do it

Most people in today's world feel that since calculators can find square roots, that children don't need to learn how to find square roots using any pencil-and-paper method. However, learning at least the "guess and check" method for finding the square root will actually help the students UNDERSTAND and remember the square root concept itself!

So even though your math book may totally dismiss the topic of finding square roots without a calculator, consider letting students learn and practice at least the "guess and check" method. Since it actually deals with the CONCEPT of square root, I would consider it as essential for students to learn.

Depending on the situation and the students, the "guess and check" method can either be performed with a simple calculator that doesn't have a square root button or with paper & pencil calculations.

Finding square roots by guess & check method

To find a decimal approximation to, say √2, first make an initial guess, then square the guess, and depending how close you got, improve your guess. Since this method involves squaring the guess (multiplying the number times itself), it uses the actual definition of square root, and so can be very helpful in teaching the concept of square root.

Example: what is square root of 20?

You can start out by noting that since √16 = 4 and √25 = 5, then √20 must be between 4 and 5.

Then make a guess for √20; let's say for example that it is 4.5. Square that, see if the result is over or under 20, and improve your guess based on that. Repeat this process until you have the desired accuracy (amount of decimals). It's that simple and can be a nice experiment for students!

Example: Find √6 to 4 decimal places

Since 2² = 4 and 3² = 9, we know that √6 is between 2 and 3. Let's guess (or estimate) that it is 2.5. Squaring that we get 2.5² = 6.25. That's too high, so we reduce our estimate a little. Let's try 2.4 next. To find the square root of 6 to four decimal places we need to repeat this process until we have five decimals, and then we will round the result.

Estimate	Square of estimate	High/low
2.4	5.76	Too low
2.45	6.0025	Too high but real close
2.449	5.997601	Too low
2.4495	6.00005025	Too high so the square root of 6 must be between 2.449 and 2.4495.
2.4493	5.99907049	Too low
2.4494	5.99956036	Too low, so the square root of 6 must be between 2.4494 and 2.4495
2.44945	5.9998053025	Too low, so the square root of 6 must be between 2.44945 and 2.4495.

This is enough iterations since we know now that √6 would be rounded to 2.4495 (and not to 2.4494).

Finding square roots using an algorithm

There is also an algorithm for square roots that resembles the long division algorithm, and it was taught in schools in days before calculators. See the example below to learn it. While learning this algorithm may not be necessary in today's world with calculators, working out some examples can be used as an exercise in basic operations for middle school students, and studying the logic behind it can be a good thinking exercise for high school students.

Example: Find √645 to one decimal place.

First group the numbers under the root in pairs from right to left, leaving either one or two digits on the left (6 in this case). For each pair of numbers you will get one digit in the square root.

To start, find a number whose square is less than or equal to the first pair or first number, and write it above the square root line (2):

Then continue this way:

Square the 2, giving 4, write that underneath the 6, and subtract. Bring down the next pair of digits.

Then double the number above the square root symbol line (highlighted), and write it down in parenthesis with an empty line next to it as shown.

Next think what single-digit number something could go on the empty line so that forty-something times something would be less than or equal to 245.
45 x 5 = 225
46 x 6 = 276, so 5 works.

	2	5
	√ 6	.45	.00
	- 4
(45)	2	45
	- 2	25
		20	00

	2	5
	√ 6	.45	.00
	- 4
(45)	2	45
	- 2	25
(50_)		20	00

	2	5	. 3
	√ 6	.45	.00
	- 4
(45)	2	45
	- 2	25
(503)		20	00

Write 5 on top of line. Calculate 5 x 45, write that below 245, subtract, bring down the next pair of digits (in this case the decimal digits 00).

Then double the number above the line (25), and write the doubled number (50) in parenthesis with an empty line next to it as indicated:

Think what single digit number something could go on the empty line so that five hundred-something times something would be less than or equal to 2000. 503 x 3 = 1509
504 x 4 = 2016, so 3 works.

	2	5	. 3
	√ 6	.45	.00	.00
	- 4
(45)	2	45
	- 2	25
(503)		20	00
	-	15	09
		4	91	00

	2	5	. 3
	√ 6	.45	.00	.00
	- 4
(45)	2	45
	- 2	25
(503)		20	00
	-	15	09
(506_)		4	91	00

	2	5	. 3	9
	√ 6	.45	.00	.00
	- 4
(45)	2	45
	- 2	25
(503)		20	00
	-	15	09
(506_)		4	91	00

Calculate 3 x 503, write that below 2000, subtract, bring down the next digits.

Then double the 'number' 253 which is above the line (ignoring the decimal point), and write the doubled number 506 in parenthesis with an empty line next to it as indicated:

5068 x 8 = 40544
5069 x 9 = 45621, which is less than 49100, so 9 works.

Thus to one decimal place, √ 645 = 25.4

Visitor comments

I vaguely recall learning the square root algorithm in K-12, but frankly, I see no value in this algorithm except as a curiosity. And I am not of the "reform" crowd. I fully believe students not be given a calculator to use until advanced algebra or pre-calculus, and then only a scientific calculator (not graphing). Do you really believe student at the K-7 level will understand how/why this algorithm works?

I was happy to see that you recommended the "estimate and check" method. This is what I also recommended to my daughter, who is now studying square roots in her home school curriculum. The "estimate and check" method is a good exercise in estimating, multiplying, and also memorizing perfect squares.

Another method, more suitable for students in an algebra class, would be to simplify the radical using the accepted method. Then find the remaining square root with an estimation method. For example, To find SQRT(1400), simplify to SQRT(100)*SQRT(14), which is equal to 10*SQRT(14). Then find SQRT(14) by an estimation method. For square roots of perfect squares, no estimation would even be needed.

One could even make the task of finding square roots into a computer programming exercise, having students write a program in javascript or some other language to use a systematic numeric method of estimating this square root via a check and guess method. Or, at the calculus level, the student could write a program that uses a Taylor Polynomial to evaluate a square root.

Michael Sakowski
Instructor of Mathematics

Howdy,

Noticed several of the comments related to using an algorithm to find the square root of a number. Some comments appeared to say that finding the result with a paper and pen vs calculator is archaic. That may be so. However, when I was in my freshman year at high school (early 70's) Herr Quinnell mentioned - as class was getting out - some of the things one can do with math - including finding square roots. So, I asked him how this was done. He showed me the algorithm method on the board.

I cannot speak to the value of generally knowing how this is used in other professions. In electronics engineering, finding square root is an integral part of design. We have parts called resistors. They aid in limiting current in circuits. These parts have wattage ratings. The value of a resistor is measured in "ohms". In a math sense this can be found by dividing volts by amperes. 10 volts divided by 0.001 ampere is a resistance of 10,000 ohms. As a square root example if I know the 10,000 ohm resistor has a rating of 0.25 watts I can calculate the maximum worst case voltage that could appear across it, before damage could occur. This is found by taking the resistance value - multiplying the wattage rating - and finding the square root. Square root of 2500 is 50. This part could withstand 50 volts.

My point - I could have calculated the result using 'artificial means'. Because somebody took the time to show me how to do square root on a chalkboard, I did not need to hunt down a calculator. By the time I would have found the calculator I've already figured out an answer. Taking the time to show students how things like square root are done has value. They may not actually put this to use later in life - but some just might.

Garth Price, CET

I was just writing another comment and somehow the computer submitted it before I was done.I must have tapped the wrong key. So let me just finish by saying that the children are new to the world and are exploring it. Calculating square roots longhand would I believe be fascinating for them and a great way to learn about other topics in math. Oh and by the way I didn't have any lessons at all on square roots until high school and then we didn't learn any way of calculating them.We were taught to factor the number under the radical and extract perfect squares leaving non-perfect squares under the radical. BECAUSE EVEN THE TEACHER DIDN"T KNOW HOW TO DO IT THE RIGHT WAY. Bye and God Bless

Robert Monroe

this is one of the very best sites I have visited for the correct process to solve a problem. You may call me arcaic but when I went to school, they taught the long division to find a square root of a number.
MOSTLY, IT TEACHES ONE TO THINK. Using a calculator is a form of pure laziness. I feel that our children think that getting the basics in school(EARLY) is arcaic. That is why when you go into the store and the bill is 16.75 and you hand the teller a twenty dollar bill, a single dollar bill and 75 cents they haven't a clue what the change should be unless the cash register tells them how much to give you. This leads to LAZY THINKING OR, NOT THINKING AT ALL.
Thank you for your time.

Rush Kerlin

I was looking at the web for the long forgotten routine for finding square roots by hand and I run into your webpage. and wanted to say that many (or all) of the criticism on the standard algorithm calling it 'archaic', 'dead end' method, etc. in favor of the Babylonian method cannot be justified. The fact of the matter is using paper and pencil to do long division or finding square roots is archaic and is a dead-end process in the 21 st Century, irrespective what routine we use, since we don't do that anymore for any practical calculations. So the issue is what should we teach to expose students to the fundamental techniques? Babylonian method is a numerical method unlike the other method, and it makes perfect sense to teach the standard routine that works for any numbers first and then other approximate numerical methods, rather than using a predictor-corrector type numerical methods saying they have applications elsewhere. If we go with the predictor-corrector type methods, one has to do an error analysis also, which is not needed with standard method since with the standard routine the correct digits are added one by one with each step (unlike the Babylonian method where the content of the digits may change through each averaging).

Best Wishes,
Karl I. Jacob
Professor, School of Polymer, Textile and Fiber Engineering
Professor, G. W. Woodruff School of Mechanical Engineering
Georgia Institute of Technology

You provided an answer to address, Finding square roots using an algorithm. I noticed that the answer provided was challenged by several people for several reasons. I would like to point out that the solution provided is THE oldest method of solving for square roots in the western world. I was described by Leonardo Picano, otherwise known as Fibonacci, in his book Liber Abaci, Chapter 14. The first edition was "written" in 1202, and the second edition was "written" in 1228. I say "written" because it was literally written by hand, as were all the copies. Johannes Gutenberg's work on the printing press didn't begin until 1436.
Leonardo learned the method from his Arabic travels around the Mediterranean sea, and the Arabs learned it from the Hindu nation around todays India. The method in the example you show, includes some modern interpretation that makes it easier to read. Leonardo also showed a geometric relationship that is related to what we understand as 'chords' today. This is a very simple, non-calculator solution to the question.

David T. Carrott, PhD

I read your suggestion for calculating square root without a calculator. I teach Math for Elementary Teachers and developmental math courses (algebra) to adults. I feel that the focus should be on understanding the number rather than an exercise in following a memorized algorithm. I suggest you have the student determine the pair of perfect squares the number falls between. For example, if finding the sqrt of 645, it falls between the sqrt of 625 which equals 25 and the sqrt of 676 which equals 26. So the sqrt of 645 has to be between 25 and 26. Where does it fall between? There are 50 numbers between 676 and 625. 645 is 20 numbers beyond 625, so 20/50 = 0.4 So the sqrt of 645 is very close to 25.4
This method provides the student with a process that improves their understanding of numbers without expecting them to memorize an algorithm, and it provides an answer to the nearest tenth.

Andrea S. Levy, Ed.D.

I'm currently a student at MCC I'm taking a course that is for Elementary Math Teachers. We are supposed to do a lesson plan so that we can teach elementary children how to use the Pythagorean theorem. I need to learn how to break down Pythagorean theorm for an elementary child. I got stuck at the square rooting part.

Read my answer to this question.

The method you show in the article is archaic. There is a MUCH more efficient algorithm. (This is the algorithm actually used behind the scenes inside a calculator when you hit the square root button.)

1. Estimate the square root to at least 1 digit.
2. Divide this estimate into the number whose square root you want to find.
3. Find the average of the quotient and the divisor. The result becomes the new estimate.

The beauty of this method is that the accuracy of the estimate grows extremely rapidly. Each cycle will essentially double the number of correct digits. From a 1-digit starting point you can get a 4-digit result in two cycles. If you know a square root already to a few digits, such as sqrt(2)=1.414, a single cycle of divide and average will give you double the digits (eight, in this case).

In addition to giving a way to find square roots by hand, this method can be used if all you have is a cheap 4-function calculator. If students can get square roots manually, they will not find square roots to be so mysterious. Also, this method is a good first example of an itterative solution of a problem.

David Chandler

This other way is called Babylonian method of guess and divide, and it truly is faster. It is also the same as you would get applying Newton's method. See for example finding the square root of 20 using 10 as the initial guess:

Guess	Divide	Find average
10	20/10 = 2	average 10 and 2 to give new guess of 6
6	20/6 = 3.333	average 3.333 and 6 gives 4.6666
4.666	20/4.666= 4.1414	average 4.666,4.1414= 4.4048
4.4048	20/4.4048=4.5454	average = 4.4700
4.4700	20/4.4700=4.4742	average = 4.4721
4.4721	20/4.4721=4.47217	average = 4.47214
This is already to 4 decimal places
4.47214	20/4.47214=4.472132	average =4.472135
4.472135	20/4.472135=4.472137	average = 4.472136

The poster asserts that the article's method is "archaic" and that the "Babylonian Method" is more efficient. At first glance, this would appear to be so, because the poster's example finds the square root of the two digit whole number 20 instead of the article's example of 645.

However, I actually worked out the article's example (square root of 645) using both methods and found that the Babylonian Method required 9 "cycles of divide and average" to arrive at the answer. Also, the Babylonian Method requires the student to perform 5 digit long division - no small feat for an elementary or middle school student. The article's method, on the other hand, only requires the student to perform one 4 step, long division problem by working out at the most a half a dozen or so 4 digit x 1 digit multiplication problems.

It is therefore reasonable to conclude that the Babylonian Method is more suitable to solve by calculator or solve by computer, while the article's method is more suitable to solve by pencil-and-paper.

Since the subject of the article was how to teach an elementary or middle school student to easily find square roots with a paper-and-pencil method, the article's "archaic" method seems to be the most fitting.

Alex

In response to Alex's post, How did it take you 9 cycles to produce 25.4 using the Babylonian Method on 645? It takes 1.5 steps if you use your guess as 25
1) 645/25 = 25.8
(25 + 25.8)/2 = 25.4

2) 645/25.4 ≈ 25.39

The Babylonian method is very effective if one already knows many perfect squares to approximate the original value. I find that students cannot follow the reasons behind the algorithm in this post, while the divide and average method seems to be more intuitive if they have worked with averages before.

Daniel

I am doubtful about teaching the long division method for extracting square roots. The Babylonian method is easier to remember and understand, and it affords just as much practice in basic arithmetic. More importantly, it has clear connections to topics such as Newton's method and recursive sequences that will be encountered in calculus and beyond. The long division method is somewhat faster for manual calculation, but it leads to no other important topics -- it is a dead end.

David

I was trained on old computer circuitry and binary hardware algorithms. The method used to calculate the root of 645 is the method used in high performance binary calculations since it only requires shift, subtract, and compare which are all single cycle/stage instructions or are diverted to a co-processor. Convert a number to binary, split it into 2 bit groups, and use the above routine. Multiply and divide require 10's to hundreds of cycles/stages and kill preformance and pipelines. It is faster to perform a square root than a divide since divide works through 1 bit per cycle/stage and square root steps through 2 bits per cycle.

Brad

what is the square root of -1?

Tamara Yardley

-1 cannot have a square root (at least, not a real one) because any two numbers with the same "sign" (+/- positive or negative), when multiplied, will equal a positive number. Try it: +2 × +2 = 4 and -2 × -2 = 4.

Since a square root of a number must equal that number when multiplied by itself. When you multiply this number by itself, and set it up as a full equation ( n * n = x ), the two factors (n and n) are either both positive or both negative since they are the same number. Therefore, their product will be positive. No real number multiplied by itself will equal a negative number, so -1 cannot have a real square root.

Blake

Square root of -1 is not a real number. It is denoted by i and called the imaginary unit. From i and its multiples we get pure imaginary numbers, such as 2i, 5.6i, -12i an so on. It leads to a whole new number system of complex numbers where numbers have a real part and an imaginary part (for example 5 + 3i or -20 - 40i). And there is a lot of fascinating mathematics done with this number system!

I was trying to find on the net the old way of doing square roots by long division. YEAH I found it. Read the responses and would disagree with many of the posters.

Finding the square of 645 is easy if you know 252 and 262 but I never memorized the squares of numbers from 1 to 30 or so, I only memorized up to 12X12 (old imperial system)

Guessing the square of 645 is around 25 is great but if you guess it's 2 then you have a larger problem ahead of you.

I see the 'other' posters are finding easier quicker ways...that is the trouble today. Let's look for an easy way with no understanding. With your method anyone with long division and simple multiplication skills can do it. The simplest solution is buy a calculator and avoid all mental skills. LOL

square root of 645 hmmmm 20
645/20 = 32.25 average of 52.25 = 26.25
645/26.25 = 24.57 average of 50.82 = 25.41

Averaging method seems to work, but it isn't teaching much division...sorta like the higher/lower on The Price is Right.

My guess of the square of 645 is 25.41....wow it works the first time, what did I learn, nothing.

Using the Averaging method, what is the square root of 9331671....my first guesstimate is 10, have fun!
9331671/10 = 933167.1 + 10 = 9331681.1/2 = 466588.55
9331671/466588.55 = 19.999785 + 466588.55 = 466607.57/2 = 233303.285
9331671/233303.285 = 39.99802 + 233303.285 = 233343.27/2 = 116671.235
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Oh yeah, these are kids in grade 3 or 4 doing long math with 8 digit numbers...so much for averaging. And what is the degree of significance since we are working with one decimal place or 3....don't want to 'average' too soon or we could lose significant digits. If we are working with billions dropping digits too soon can make a HUGE difference.

adrian

I'm a layman who came to the site via a Google search on "how to calculate a square root." I read the presentation, then looked at the responses. I must say that I was dismayed at the comment offered up by Andrea S. Levy, Ed.D., where she suggested that memorizing an algorithm is less desirable than understanding a number.

I presently work as a technical writer for a firm that writes credit union banking software. Understanding all the algorithms used in the financial world is utterly essential for us to do what we do. In fact, one of the calculations we use to determine the amortization of a consumer loan with fees in a given time period is strikingly similar to your square root presentation. The calculation must be written by the software engineer for the machine, so it does ultimately reside in the mind of a human being. If the engineer doesn't know the algorithm, thousands of consumers will bear the consequences. I suggest that memorization is simply another tool in the box. Use it when its appropriate.

Best regards,
Michael Kelly
Newbury Park, Ca.

The last commenter on the page (Adrian) said that she never learned the squares from 1 to 30. This brings to mind a trick I recently learned for finding squares close to 50. Start with the square of 50, 2500, add 100 times the distance between 50 and the number, and then add the square of the distance of 50 and the number. For instance, 43² = 2500 - 700 + 49 = 1849. This comes from the simple FOIL identity (50 + x)² = 2500 - 100x + x². In this identity, x is the distance between 50 and the number. If the number is 43 (as in my example), x is -7. If the number is 54, x is 4. So if you memorize your squares from 1 to 25, you get the squares of 26 through 75 "for free".

If the idea of memorizing the squares of 1 to 25 seems daunting, it's not. A few weeks ago, before knowing this trick, I knew just up to 13 offhand, with a few others scattered here and there. I drew up a table in Excel listing numbers 1 to 25 side by side with their squares, printed it out and put it on the wall of my cubicle. The squares I don't have memorized in those first 25 I can now get in a few seconds (for instance, for the square of 23 I am still counting up from 20 squared: 400, 441, 484, *529*). Even with not quite knowing them all I can find squares from 1 to 75 in under 10 seconds (thought process for finding 73 squared offhand: "73 is 23 greater than 50. What's 23 squared again? 400, 441, 484, 529! 2500 + 2300 + 529 = 5329. Done!")

David Levy