Find Square Root by Hand without Calculator
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- เผยแพร่เมื่อ 27 ก.ย. 2024
- Learn how to find the square root of a number by hand approximated to at least two decimal places. In this video we approximate the square root of 38 out to two decimal places without a calculator.
Video Link: tinyurl.com/mo...
#squareroots
#byhand
#withoutcalculator
My high school teacher insisted there was no way you could do square roots by hand. I insisted there was a way and she challenged me to come up with a method. I never figured it out, but I knew there HAD to be a way. How else could people do them without calculators for all those centuries?
You've shown the way! :D
Same here-it seems like most math teachers like to keep all the cool and interesting math to themselves. My newest math tutorial video shows a *much easier* method for approximating square roots intended to replace this one-I hope it helps you! th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
Almost as bad as Math teachers who don’t know there is a quick way to tell if 7 evenly goes into a number. They show 2, 3, 4, 5, 6, 9, 8, 10, but seem to leave out 7 for some reason. 😡
I.e. Double the last digit and subtract that from the remaining digits. Repeat until left with -7, 0, or 7 which means the original number is evenly divisible by 7.
Somehow *good* math textbooks from 70 years taught it but modern ones don’t. 😳
How did she say it was done before the advent of calculators or computers?
how else would they program the calculator?
you should find her on linkedin or facebook and send her this video
A true teacher!
You don't presume we know but take the trouble to take us through all (not some) of the logical necessary steps to the answer.
Thank you, sir!
This is the sort of comment that makes it all worth it...thank you, truly, and I am happy to help. Good luck to you. -Mike
@@AthenianStranger I dropped out of school, but I’m currently learning how to do calculus by myself, thank you for everything you do❤
I was once shown this by a math teacher when I was 13 and have been trying to remember it ever since. Thank you for this video.
Glad to hear it…I actually no longer use this method and will soon upload a much, much easier method which I use to teach students aged 9-12 how to do the whole thing in their head and it is correct to the tenths place.
Keep an eye out for that video. It is so much easier that it makes me cringe to see this video has so many views!
Been about 45 years ago for me, we learned it before the advent of personal use calculators.
@@AthenianStranger I think this video fulfills a different purpose than your new video, because this method makes it possible to calculate square roots to an arbitrary precision
The in your head method is the only useful one in our day and age, because if I need precise square roots I can just use a calculator
My Dad showed it to me when I was about 10. Similarly, I've thought about it from time to time. I only remembered the additional two zeros but forgot the important part of doubling and multiplying by the magic number. Thank you for this video!
HHHGeorge
I hope this not an example of how you have spent your life. Just DO it.
This is how I had learnt to find square roots some 45 years ago in Indian schools. The format and layout were slightly different but the flow was the same. Calculators weren't readily available then. Thanks for this demo.
Did they also teach the theory of this method, the "why" behind the "howto"?
i have never known a time without cellphones and calculators. i have caried a super computer in my pocket, more powerful than the ones used in the moon landings for 20 years.
@@Tubemanjac I need that too. Otherwise this is just a memorized process that makes no sense to me.
🙂 Yes. In Romania we learned this in school as well. And our method was simpler. Same flow, easier logic.
This was first documented by Madhava of Sangamagrama in the 1300s. He also discovered the first recorded formula for calculating pi using infinite series, which is the basis of modern calculus, centuries before Newton or Lebnitz.
Thanks for being the one video that actually explains this process instead of just a “trick” to do rough estimates.
You’re welcome 😊
thats exactly what i was thinking! great video and great explication
@@toplel_96 Me too!
I thought this was my comment for a second 🤦♂️
Actually, the failure of this video is that it only teaches mechanics and does NOT explain why it works. If you can explain why what this video shows works, then you will really understand what the square-root means as a function and will understand a lot more math all at once, including interpolation and principles of calculus. Learning is not about memorizing algorithms, it should be about understanding how they work.
Fortunately, my 3rd grade math teacher taught us this. It has turned out useful on many occasions.
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
@@AthenianStranger The second one is called the Bakhshali method, for anyone interested
You learned square roots in 3rd grade??
What can I really say to explain how grateful I am for your clear explanation? Understanding mathematics does so much for self-confidence not to mention giving you the ability to breakdown important concepts and it is so often so difficult to make the process of learning clear and not something that makes you doubt your abilities. Thank-you very much.
This may be among the kindest and most personally affective compliments I have ever received on this channel. Thank you so much for your elaboration regarding the importance (and rarity) of clear, simple, and direct explanations of math. You have hit upon my central goal as a math teacher. Though I, too, must very often fail in this regard, I try to teach math to “my younger self” the way I wish it had been explained. Perhaps like you, I went from K-12, four years of college loaded with advanced mathematics, and two years of graduate school and managed to emerge without a single coherent dusting of true understanding regarding virtually all mathematics. Then, I became a teacher and never wanted to teach mathematics because I had this secret-my degrees stated that I should by all accounts have the ability to teach everything from rudimentary addition of single digit numbers through advanced calculus but in reality I didn’t have a clue about any of it. Then, one year all my school’s math teachers left-all of them-so my principal asked me if I would be willing to teach Algebra 1 (because in Texas that is the only high school math course with an end of course state assessment). I accepted this challenge and went home and wept into my pillow for trying to impress my boss but it was too late, so here’s what I did: I went on eBay and bought all my old math textbooks in their exact editions from the 1990’s (I was born in ‘86) and then I paid for this online test prep service for the math exam teachers are required to take and pass with a score of 85 or higher. My first attempt at the test earned me a 40 and I went back to weeping in the pillow. Then I just came up with the “keep it simple stupid” plan and worked 100 math problems every day starting with early childhood math (for these I worked 200-300 problems per day) and slowly worked every single problem in every single book all the way through calculus-every day, 7 days a week, for like 6-7 weeks straight with no breaks at all, 12-18 hours a day. The end of summer vacation approached and my principal said I had to take the certification test or he would have to find another solution. Ugh!!!! Pillow weeping ensued, then I just stood up, registered for the first available test which happened to be offered the next day far away on the other side of Texas so I paid for it, hopped in my car, drove straight through the night, and arrived at the testing center just in time. The test was 100 questions and I was given 5 hours. The room was cramped and packed with people so it was hot and humid. I took exactly 4 hours, 59 minutes, and 57 seconds taking the test and obsessively checking over all my work and hit “Submit and End Exam” with literally 3 seconds on the clock. I immediately called my principal and told him I failed for sure. It was the hardest test of my life-NOT A CHANCE that I passed with a 70 much less the required 85. He told me to calm down and I drove home and waited the mandatory three days for the score: I opened the email, clicked the link, logged in, and saw a word I would never have believed I’d see: “PASSED”. I passed the test by just two questions but that was irrelevant-I PASSED and was at that moment forever transformed from a history teacher into a math teacher, spawning the genesis of this channel, and now I have taught every grade mathematics from 7-12. Every time I teach, I learn right along with the kids. I think the kids and perhaps the viewers of this channel sense that-it’s why I always very slowly work each problem and explain every step-and why I use my infamous clear ruler 📏-“straight lines make straight minds.” I am just like all my viewers: Trying to learn math. The journey’s duration is equal to all our respective lifespans and never reaches a conclusion which is what makes this math adventure so interesting and forever awaiting that next newly discovered or mastered skill or rule or hidden internal logical beauty. So thank you for your compliment and for sharing this infinite quest with me 😊. Best wishes and glad tidings, Mike
That’s not a clear explanation at all!
Great explanation, not really a method for fast calculation but it makes the process so easy to understand
You should also do a video on how to do logarithms by hand. Its an equally interesting process... and gave me a ton of respect for what Napier and Briggs had to go through to make their first log tables. Yikes!
*Very good idea*!
@@AthenianStranger I hope you do go over log by hand; I absolutely abhor the prospect of being forced to use a calculator without understanding the principles at hand, and my senior year, I think ill be in a pre-calc class, something I presume will touch on logarithmic junk.
@@AthenianStranger - Yes! I would be very interested in logarithms.
@@AthenianStranger Yes, please do so! Other YT channels have shown that this can be done, e.g. Numberphile, but not made any effort to actually teach the technique. At least, no where near as clear as I believe you could. You have a desire to teach what you know and not just show off what you know
If I might suggest, stick to base 10, we know the log(10) of 10 is 1 and the log(10) of 100 is 2. We know that any number between 10 and 100 will have a log between 1 and 2. How could we calculate the log of say, 20 or 50 or 66? No more than 2 or 3 decimal places and I think we'd get the idea. You know, lather; rinse and repeat.
I believe you'd have to always calculate an extra digit to determine approximation, for example, if you just calculated the first decimal, you'd get 6.1, however, in the video you can see that this would approximate to 6.2, so calculating that one extra digit every time should increase the consistency of success in this method.
I just recorded a much simpler method for approximating square roots. Here’s the link: th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
Unless the digit is 4, in which case you'd need to keep going
@@stoppernz229no
This neat method is based on the binomial coefficients. For square root, it's derived from a^2+2ab+a^2 as such: a^2 + b(2a+b). A similar but more complex method gives cube root: a^3 + b(3a^2+3ab+b^2). For fourth root, it's a^4 + b(4a^3+6a^2b+4ab^2+b^3). Fifth root is a^5 + b(5a^4+10a^3*b+10a^2*b^2+5ab^3+b^4), and so forth. You would bring down groups of three 0s, four 0s, and five 0s, respectively.
It's doable by hand for cube root, but rapidly grows to involve gigantic multiplicands when used for higher roots. Nonetheless, it works.
Note, you don't have to actually 'double' the number to get the next number in sequence: the pattern shown is this: let's say we have 12_ at the start. You find it to be the number 1. Hence, 121. The next number will be 121+(number you found=1)_. Thus (121+1)_ = 122_. We found in the video that the missing number was 6. So the next number in the sequence will be 1226+(number you found, thus 6)_ = 1332_, etc.
1232_
Wow.
*1232_, but yes.
It is 1232, 1332 is wrong.
Does it work for all equations?
A very good explanation - thank you.
I would add that it would be better to show the next decimal place, too. I was taught that, if the answer was to 2 decimal places, you had to work out the third dp. If that figure is 5 - 9, you have to round up.
IF you had to round up, then how would YOU define Pi..3.14157... to a never ending line of numbers. there is NOTHING To" round up". The answer is is a never ending series of numbers.
@@guysabol8743 um... that's the whole point of "rounding". If you wanted to round 3.14157 to the nearest ten-thousandth, it would be 3.1416. Nearest thousandth would be 3.142. Nearest hundredth would be 3.14 (look familiar?)
Cool method!
I like developing a Taylor series around the value (it's simpler than most people think, but one does need to know derivatives). I can get to 2 or 3 digits behind the decimal point quite quickly.
Can you explain it further Rony I’m interested
For the example of sqrt(38):
Re-express the expression to be solved as:
sqrt(38)= sqrt(36+2)= 6*sqrt(1+1/18)
We factored the closest perfect square out of the argument of the square root.
Then use the Taylor series of sqrt(1+x), with x=1/18.
The Taylor series is:
sqrt(1+x)= 1+x/2-(x**2)/8+...
So in our example:
sqrt(38) = 6*sqrt(1+1/18)
sqrt(38) = 6*[1+(1/2)*(1/18) +...]
sqrt(38) = 6*[1+1/36+...]
sqrt(38) ~ 6 +1/6
The next term subtracts a less significant amount to correct the result downwards, then the one after that adds an even less significant amount to correct the result upwards, then downwards again, then upwards again, and so on.
You can look at the sizes of successive terms to determine whether you care about continuing on to the next term.
For it to work, the absolute value of x has to be less than 1.
It works better (you need fewer terms for a given desired precision) as the absolute value of x becomes smaller compared to 1. For larger values of |x|, it's not as good of a method. It therefore works better when the argument of the square root is close to a perfect square.
If you are close enough to a perfect square to only need a first order correction in x, then you could use the following formula:
sqrt(N^2 + x) ~ N + (1/2)(x/N)
Taylor series of sqrt(1+x):
www.wolframalpha.com/input/?i=taylor+series+of+sqrt%281%2Bx%29
@@_Longwinded it's a concept introduced late into calc 2, look for taylor series approximations in the comprehensive calc 2 guide on youtube. even if you don't quite understand it it's nice to know how it's performed
Yeah, Taylor series centered at the nearest perfect square number was the first thing to come to my mind, although as for the square root one can use the so called Babylonian method or simply write down a quadratic equation x^2 = a and solve it numerially using the Newton's method (which can be further generalized for an arbitrary nth root problem)
you know he's legit when he pulls out the protractor to finish up that square root
I’m obsessed.
Thank you so much. I understood nothing on my online class when teacher was explaining it. You helped me a lot
Simple, but not precise method is optionally:
1. Estimate the root as well as possible - Lets take the same example (y = 38) and first estimation a = 6
2. Compute the next value as b =(y + a^2)/(2a) e.g. b = (38 + 36)/12 = 6.166666 what is quite good result (delta
It is still better just using a Taylor expansion centered at the closest squared number to your desired value:
√x ≈ √xo + (x-xo)/(2√xo)
Use xo = 36 since this the closest number to x=38 with exact square root:
√38 ≈ √36 + (38-36)/(2√36)
√38 ≈ 6 + 2/12
√38 ≈ 6 + 1/6
√38 ≈ 37/6 = 6,16…
With this method you will always get some excess in the aproximation, this is: √38
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
This is way better, and more intuitive, than the method that I was taught in school.
Thank you.
I like that you use a transparent protractor as your straightedge. Me too! For only a couple of decimal places, seems like it would be simpler to just guess and check.
AWESOME! Thank you. Its just what I have been looking for to do Fire Dept Hydraulic water calculations for Gallons per minute coming out of the fire hose. Without a calculator. Formula is: (GPM=29.7 x d squared x square root of nozzle pressure) (d squared is the nozzle tip size. ie 1 ", 2", etc) (nozzle pressure can be 80psi or 50psi, etc)
I can get 1 decimal pretty quick, which I think is neat if someone wants you to do it on the spot.
for 38 you take the difference between the closest perfect squares 36 and 49. the difference is 13 then you take the remainder from the smaller square so 38-36 which equal 2
and you divide the remainder by the difference so 2/13 = .15.... and add that you 6 and you got a 1 decimal approx and its within a .2% error, not too shabby IMO
obviously workes better for some numbers rather than others
FORMULA:
X = the square root answer
Y = the square closest to the number
Z = the square root number
Difference = the Difference you get after Subtracting Z from Y
√Z = X . _____
Z - Y = Difference
Difference / X * 2 + X = X
Example:
√26 = 5 . ______
26 - 25 = 1
1 / 5 * 2 = 1 / 10 = 0.1 + 5 = 5.1
√26 = 5.1
For readability, every time you write "difference" you should replace it with a variable named "D" described as the difference of Z from Y
I've been curious about this for decades! Great instruction, sir. Thanks!
Thanks so much for the great tutorial on square roots! Extremely clear and very helpful!
Glad it was helpful!
This elementary mathematics was taught to our generation in 8th grade math class. Slip sticks were common also for generating educated math replies. H/P made the best slide rules available at the time.
You've really explained it to my satisfaction.
Not the petty petty tricks that ends up confusing you more.
My Brain is the size of a walnut
Thank you! I am 67 and had never had that explained to me ! ( Girls used to be told " you can't do this. ") You are very clear, knowledgeable and good at explaining math. Thank you.
You are so welcome!
Yes, I recalled this approach from my school days (many many years ago). Applying this method to binary is truly magical! It's rather easy to create a square-root engine in digital logic! The only choices are between 0 and 1.
Thx for that. Just went through example after converting the radicand 28 to binary. Very cool!!
I'm 48 years old and never seen someone do this... Thank you.
i notice that the double is equal to the sum of the previous two numbers that were multiplied. So, 122 = 121 + 1. And 1232 = 1226 + 6. Could be coincidence of course, but if it's not then that's a slightly easier shortcut.
You are correct. I found another tutorial that did it this "addition" way, rather than "doubling". It appears you get the same numbers either way.
I learned how to do this many years ago when I was in school for drafting. Long since forgotten how to do it. Hopefully this refresher will stay with me, for no other reason than to know the skill.
I just recorded a much simpler method for approximating square roots. Here’s the link: th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
youre a lifesaver, unit final is tomorrow and i had no clue what i was doing
Fascinating thank you! It’s crazy we were never taught this in grade school
This is one of the coolest mathematical tricks I’ve seen, and it makes intuitive sense. Thank you!
Glad you liked it! I now use a much simpler method-please give it a look 👀 : th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
thank you so much! this video deserve more views
Thank you too!
Nice work. I learned how to do square roots on paper more than 60 years ago but had forgotten how. Thank you for reminding me how it is done.
to be totally confident on the two decimal places, would you need to check on the third decimal place, in case it were to round up?
That is a neat method if you can’t find your slide rule, which is how I would typically do it for only three significant figures (just transfer from B to C scale). The nice thing is that this method expands farther than three significant figures, a slide rule can’t really do much more than three.
Why? Im 82. I took square roots by hand for years before calculators could be held in your hand and could be afforded by school kids. But to develp this skill today is VERY low priority. I like your videos even though they are quite verbose. But your recent videos on doing calculator worthy tasks without a calculator seems crazy. I guess your slide-rule series will be next!
My father showed me this 40 years ago - and also a similar method for cube roots (you bring down 3 digits at a time, and the running calculations are more complicated).
When I was I kid I asked my math teacher to explain to me how can I calculate the square root of a number by hand. She was unable to give me an answer. After a few decades thanks to you now I know how to do it :) Thanks!
I just recorded a much simpler method for approximating square roots. Here’s the link: th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
I invented my own method several decades ago which initialy started with trial and error and then I refined it into more of a system which requires two calculations per decimal place.
One thing that I have observed is that if a number is not a perfect square then its square root has an infinite number of significant digits. In other words, you can go on forever making the square root more exact. I also want to show you something weird, but it ends up working anyways.
38 - 36 --> R = 6; D00 = 200
200 - 120 * 0 --> R = 60; D00 = 20000 ; Whoops! I picked the wrong digit. I should have picked "1" instead, but there is a reason for this stunt.
20000 - 120[16] * [16] = 544 --> R = 60[16]; D00 = 54400; This is exactly the same remainder that you had after 3 digits; Therefore, we are now back on track. If we were to stop here then you take the quasi-number 60[16] and do a conversion like this --> [16] is too high a digit for base 10. So you subtract 10 from [16] and make that a "6" and carry a "1" (1/10 of the 10 you just subtracted) to the next place on the left. That then causes the "0" to become a "1", and you now have "616" as you are supposed to have.
To calculate 120[16] * [16], I went [16] * [16] = 256; leave the "6" and carry the "25"; I then went 0 * [16] and added "25" to get "25"; leave the "5" and carry the "2"; I then went 2 * [16] + "2" to get "34"; leave the "4" and carry the "3"; I then went 1*[16] + "3" to get "19"; no need to carry since we are at the end; So, we have "19456"; 20000 - 19456 = 544.
Basically, this worked because all numbers in base 10 are the result of plugging the value "10" into a regular polynomial (a polynomial in which all the powers are integers) in which all the constants are integers, too.
You're right about the significant digits thing; in fact, it's possible to prove that perfect squares are the only numbers that can have rational square roots. If the root of some number x can be expressed as a ratio of integers m/n, you can deduce almost immediately that m and n must both have a factor of x. Therefore, the ratio m/n must reduce to an integer in order to avoid an infinite loop, which means that x is a perfect square.
@@Cornix94 I assume you mean to say, "perfect squares are the only integers that can have rational square roots." Small but important difference.
I watched a few videos trying to understand this and yours was the only one that made sense!! Thank you so much! Appreciate you taking it step by step.
Thank you for sharing your kind words.
I just recorded a much simpler method for approximating square roots. Here’s the link: th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
I hope you find it useful and easier to do using mental math or, at most, a scrap of paper. It is a version of what is typically called the Babylonian Method but simplified to a 1 step process because I had to find a way to make this possible for my students-ages 9-12-to do and understand.
Thank you for watching my videos and supporting this channel.
Sincerely,
Mike
why does this work? it's cool that you can do square roots by hand, but why does this work?
The only and the most amazing video out there on TH-cam that actually explains it.
Thank you for your compliment but I no longer use this method and need to post a new video showing a WAY easier method.
@@AthenianStranger very excited to check it out, pls upload as fast as u could, and thx.
This was the best video on square roots that I found, and the easiest for me to keep up with!
Excellent Video!
Did not see the justification for multiplying by 2 at each stage. The process you presented certainly produces great estimate. Maybe it is something to do with Newton's Method. Thanks for demonstration.
Simple… the square root of N is equal to X + Y DIVIDED BY THE SUM OF X + square root •••
where N is equal to X^2 + Y
The formula for determining the square root is (10X+Y)(10X+Y) which equals 100XX+20XY+YY. 10X is the number already determined and Y is the next digit being determined. X is times 10 because the digits on the left are 10 times that of the new digit being determined as a place holder. After the first digit is determined the 100XX is no longer used and you have left Y(20X+Y) for determining the next digits ad infinitum. He is multiplying the number already determined(X) by 20 then adding the new digit(Y) before multiplying by the new digit Y. The space he is leaving to put the new digit is the where the zero is of the 20. This method is more than an estimate as it is accurate ad infinitum.
This process gives exact decimals and is not based on Newton's method (at least not directly). The 2 comes from the expansion of a binomial square. At each step, consider x as the current estimate (6, 6.1 or 6.16). Then we have 38 = (x + c)^2, where c is the remaining decimals. (x+c)^2 = x^2 + 2xc + c^2. So the remainder is r = 2xc+c^2 = (2x + c)c. This corresponds to the xxx_._
It doesn't produce an estimate, but calculates the real value.
The justification for the two is probably (it's just a guess of mine):
(a+b)^2 = a^2 + 2b + b^2
So, when you square 7.2, it's (7 + 0.2)^2 = (49 + 0.04 + 2 * 1.4) = 50.44
Now, when you want the squareroot, you have to do the inverse process, so you have to get rid of the 2 somehow.
(a+b)²=a²+2ab+b²
The "2" in the "2ab" term is where the multiplication by 2 comes from.
For example, suppose you are trying to find the square root of 1849. You can tell right away that it is between 40 and 50. So, let (a+b)²=1849, and let a=40. Then we try to find b.
1849=(40+b)²=1600+2ab+b²
Therefore, 249=80b+b²
which we can rewrite as 249=(80+b)b
Notice that we *doubled* 40 to get the 80 in this equation.
Then we can use guess-and-check to find b=3
and therefore (a+b)=40+3=43. Therefore, the square root of 1849 is 43.
I wonder what is the benefit if we are looking for an estimate.
We know that the answer is between 6 and 7 (because 38 is between 36 and 49).The distance between two perfect squares is 13, of which the distance of 2 (38-36) is less than 1/5, so the estimate will be
Your solution is an estimate - the solution you get from following the algorithm in the video is accurate (to whatever precision you want). You're assuming that the square root function is linear, which it _isn't_ . The estimate can still be useful for quick mental calculations with small numbers, of course. Indeed, linear approximations were always common in computing for things that don't particularly need to be accurate - especially in things like real-time 3D graphics. Of course, more modern methods are both fast _and_ accurate - for a CPU/GPU :P
This was so helpful 🥰 was literally crying out of stress since I had a math long quiz and I had no time to study 🥺 you explained it very clearly and this will help me so much 🥰
I've learned alot about squareroots on the way to this video. Here I learned how to really do it!
Thanks!
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
It may be good if u have the time and can understand it, but I got the same answer in a few seconds mentally and anyone can do it. The square root of 38 is 6, remainder 2/6. We can halve that to 1/6 (or double the divisor to 2/12) which is .16 as 6 goes into 100 16 times. We can even take it a step further in that there is 4 remaining from 100, so the next digit is 4, thus 6.164
This is a nice albeit somewhat slow approach to more precision, but you can get nearly the same precision you worked out instantly with the first order Taylor expansion at the nearest perfect square. √38 ≈ 6 + (38-6²)/(2×6) = 6.16666...
this process can be done to infinity
This was first documented by Madhava of Sangamagrama in Kerala, India, in the 1300s. He also discovered the first recorded formula for calculating pi using infinite series, which is the basis of modern calculus, centuries before Newton or Lebnitz.
At least in this case, it would be quite easy to find the answer with a binary-style search, anyway. This weird division stuff seems a lot harder than just "square 6.10, 38 is bigger than result -> square 6.20, smaller -> square 6.15, bigger -> square 6.17, smaller -> square 6.16" and hey presto we've got 2 digits... and if you want to be sure of which way it rounds, you can check 6.165^2 and see that it's bigger than 38, therefore 6.16 < sqrt < 6.165, therefore at two decimal places it rounds to 6.16.
IOW, you can get closer and closer by looking where you "expect" the square root to be based on squares you do know (f.e. 38 is between the squares 36 and 49 so the root has to be between 6 and 7, or even 150 is between the squares 100 and 256) and simply seeing what square you get out. You can even optimize it using your intelligence; rather than doing a true binary search starting at 6.5 (the midpoint of the search range), you can start at 6.1 or 6.25 or something since you know 38 is much closer to 36 than it is to 49.
Agreed 👍 I now use a much simpler method-please give it a look 👀 : th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
A really quick-and-dirty way to approximate the square root of b is this: Let a be the biggest perfect square less than b. Then sqrt(b) = sqrt(a) + (b-a)/(2*sqrt(a)).
For example, to approximate sqrt(107), you have a = 100 and sqrt(107) = 10 + (107-100)/(2*10), which gives 10.35. The answer is approximately 10.344, so it's really accurate.
Thank you for the information.
66 y/o here. We learned how to get square roots by hand, but it wasn't like that. We did end up with a stack of lines and numbers, but it seems like there was more guesswork. This way seems more direct.
I now use a much more direct and simpler method-please give it a look 👀 : th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
Did it works for every single number?? Oh my god this is amazing! I can't wait to show my friends and teach them
I kind of get the feeling that this should be taught to all grade schoolers by the time they reach the 5th or 6th grade. It is a shame that it isn’t because the Board of Education has failed us.
Preach brother. I couldn't agree more. I teach 7th-12th graders and make them approximate everything possible without a calculator. I don't use the method shown in the video anymore and need to make a follow-up. At first, most kids get frustrated with all the pencil math but they quickly become empowered by their ability to do "big boy" math without a machine.
This explains better than the first vid I saw on someone's channel.
Hi there, thank you for your nice compliments-but even this video isn’t showing the easiest method: I am going to make a video explaining this in a completely different way…honestly I find the method I show in this video ridiculously hard and I don’t use or teach this method anymore. Keep an eye out for a new video about how to approximate square roots and hopefully it will be more clear. Thank you, Mike
Perfect, now I know what to do, if I'm lost in the wilderness, and needs to draw the square root out of something!
So I've been doing it entirely wrong but getting the right answer much quicker than the right way. My self worked out algorithm is what I call guess and go. I just go with the largest perfect square and then start making educated guesses until I have it to the decimal places needed.
I love your dedication and how you decided to do it at 5:39 am😆
This is really cool I just tried it a few times and double checked with my caculator and I was correct. Thank you.
Wow! I remember this from high school. We were not allowed to use those new things called calculators, not even slide rules. Everything by hand. Over 6 decades ago.
Ty! I am a 7th grader wanting to learn lol
I AM GRADE 8!!!🥲😅
👌
Tܴܰhis is a simple method to estimate the square root rounded up to 0.1. We know that SQRT(38) should be between 6 and 7, i.e. SQRT(38) = 6 + some value x, Then take square of both parts of equation: 38 = 36 +12x + x^2. Since x < 1, its square will be even smaller value, and for the estimate we can dispose of it. 38 apprx = 36 + 12x, from here 12x=2, i.e x = 1/6, appr 0.16. Adding this to 6 gives us an estimate: 6.16, rounded up makes 6.2. If you need an estimate up to 0.01, then calculate the next iteration: SQRT(38) = (6+1/6) + y, 38 = (37/6)^2 + 14y/6 + y^2, etc.
Nice informative video... I really appreciate your effort...
But there are another method so we can calculate sqauare root in no time...such as
Suppose we take 38 as number which is imperfect sqaure ...... firstly find closest interger whose sqaure is close to 38 which is 6 (their square 36 close to 38).Hence first value will 6 then rest of decimal value can calculated in following easy way
38-36=2 and multiply closest interger(6) with 2 giving 12 then divide 2 by 12=0.16
Final answer=6.16....by this same method we can calculate of any imperfect square number very easily.
I now use a much simpler method-please give it a look 👀 : th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
@@AthenianStranger yes ofcourse... thanks for this
I am not sure if I am more impressed by the maths.... or the lovely handwriting!
There is also a very similar way to do cube root by hand. I learned it, but it was too tedious, and I had no need for it anyway, so I only did it the day that I learned how to do it.
A step back in time moment for me :D. I haven't done this since school back in the 70's!
So, if you'd like to stop after a single decimal digit, this wolud produce 6.1 instead of 6.2, because it couldn't account for correct rounding.
And if you want several extra digits, it soon becomes unpractical.
The niche for application seems quite narrow. Better just keep to log ruler as a backup for calculator)
i feel like the better intro would be "hello. welcome to hogwarts. this is mathematical wizardry 101"
And what happens when the number is 12 digits?
You start with however many digits you can pull a square root from (pairing two digits at a time to the left of the decimal point. If you have a number like 123456789012, you break it up to 12,34,56,78,90,12. As 3*3 = 9 and 4*4 = 10, the first digit is going to be 3. Repeat the process, bringing the next two digits down, instead of 00. The decimal point in the answer goes where after the digit you get when you bring down the pair of digits immediately to the left of the decimal point. To check, the number of decimal points in the answer is the number of pairs of digits before the decimal in the original number. If there is an odd number of digits before the decimal in the original, just put a zero in front of the front digit. For instance, treat 147 as 01,47.
Your paper magically summons a calculator :)
A year later. Just pair up every 2 digits from right to left. The first number will always be solo or part of a pair (your case).
Youre basically drilling it down drilling it down and drilling it down until you hit the acceptable root assuming you have to stop at some point because the number maybe irrational at least within human calculational ability.
This is easily the coolest thing I’ve learned today. Thank you!!! 😊
It would be interesting to solve this also by the Babylonian method, and assess whether that takes more or less hand computation. Or perhaps work it until we feel we've done an equivalent amount of hand computation and see how many digits we got with that effort. When I start with an initial guess of 6, the Babylonian algorithm gives me 6.167 after just one iteration. Granted, though, that was using an exact division of 38 by 6. So, 36/6 = 6.333, (6.333 + 6)/2 = 6.167. It converges very quickly, but I don't have a good feel for the work involved with doing that by hand.
Ask and you shall receive: I wholeheartedly agree so a few months ago I made an updated version of square roots by hand using a modified version of the Babylonian Method (way, way easier):
New Video: Find Square Roots w/o Calculator by Hand Quickly & Easily
th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
Wow i got this
Some 40 years ago, there was an algorithm going around that used a four function calculator. It used a quite simple sequence of repeated operations that added a digit with each 'round'. I don't recall it now.
Thank you for your comment - while I'm not sure what method you're referring to, I have just recently uploaded a video showing a *much* simpler and intuitive method than the tedious method shown here. Please give the new video a glance and let me know if it's the one you learned 40 years ago: th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
@@AthenianStranger It was some successive operations possibly with an add, subtract and divide, but I can't recall anything. A the time I had a small Omron calculator. Since I started with vacuum tubes in grade school, I called it a one (vacuum) tube calculator. It had a blue-green, seven segment vacuum-fluorescent display. ;)
@@AthenianStranger I looked at that one. Since I can't recall much, it's difficult to say, definitively, but it seemed fairly simple without needing any note taking, but I vaguely recall entering what possibly was a new guess digit each iteration. . . Sorry.
Dude taugth me more about math than a school in the past 5 years
And here’s another 5 years: I now use a much simpler method-please give it a look 👀 : th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
Some of comments raised the lack of justification of doubling the trial entry. Please next U tube on complete the square in Algebra,how it applies to to Square Root Algorithm. Greg
I found two methods since making this video which I use now instead and honestly I can’t even remember how I did this method (I’d have to rewatch my own video). The first method is called the Newton-Raphson method, check it out, it will produce correct decimals depending on how many times you iterate the steps. The second method is waaaaay faster and pretty accurate-Example: sqrt(13) ~ (13 + 9)/{2 * sqrt(9)}. The way I came up with the 9 is because it is the largest perfect square which is less than the number (13) we are approximating the square root of. In this example we approximate the sqrt(13) ~ 22/6 ~ 3.67. Actual sqrt(13) ~ 3.61. Pretty darn close and you can do that second method in your head.
due to how complex this method is, i think this would drive a mathematician insane
so i tried it with 1101.64 and here's how it went:
1. i had to *estimate* which square is the largest without going over, most commonly done with *trial and error* (i got 33)
2. oh right you have to divide it into groups of 2 digits (__ . __ is ok, and not _ _._ _)
3. write the estimated number to the left of the √ symbol and above (the last digit of that number should be uhhhhh just write that placed down like normal long division)
4. the decimal i wanted to square root had digits after the decimal point, so copy that down (but if it has 1 or 0, then just write zeros instead) (so i have "≤ 1264"
5. uhhhhhhhhh 2:12
yeah i have no idea how to explain this part (i write 1 here)
6. .............
…………………………………………
and you just repeat step 5 and 6
liked and subbed, ive been looking for this all over the net, thank you
Doesn't make any sense
Agree completely.
Dang... I would do that a lot simpler: 6 squared is 36... 6,2 squared (pretty doable by hand) is >38, so 6,1 it is.... 6,15 sq. (still doable by hand 38, fails. 6,165 sq >38, fails, so 6,164 it must be. And so on and so on.
Estimate the next decimal (in case of doubt start with the middle of the next decimal range, a.k.a. "5") and go from there.
Easy method: 6 is a good estimate, but too low. Hence 38/6 is another good esitmate but too high. So their average 1/2(6+38/6) = 37/6 is a very good estimate with a nice advantage that it is a rational number. And pretty accurate: it is approximately 6.1667 whereas to exact answer is 6.1644, so the error is 0.04%.
If you need a better estimte repeat the process using 37/6 as starting point. 1/2(37/6 + 6*38/37) = 2737/444, which has six correct decimals.
If the challenge is approximation, I would simply use the derivative (2x).
38… it is more than 36 and less than 49… first digit 6…
The derivative is 2x. In other words the slope of the curve at the value of 6 is 2x6=12…
Take the remainder 38-36=2
And divide it by the slope
2/12= 0.1666
Add this to the first digit…
6.1666…
This works like magic with higher numbers… 8460
90x90=8100
90x2=180
8460-8100=360
360/180=2
90+2=92
While exact answer is 91.978
I was an English major who always sucked at math. Always. I’m old now. I have no idea why this showed up in my feed, but I really enjoyed it. Almost enough to forgive Mr. Cash, my sophomore Algebra teacher, for failing me! 😎
Glad you found this helpful; however, I must admit this method is very tedious and I long ago abandoned it for the method shown in this brand new video I just released showing a method so simple I can (and do) teach it to 9 year olds who can compute most of the roots in their heads, check it out: th-cam.com/video/xIpoxUPfuRY/w-d-xo.html
Or, you could do the algorithm that is used by all calculators.
Take the next least perfect square and add 1. divide into the number and then add then divide by 2. Take that number and repeat.
so for 38 the next least perfect square is 6. Add 1 and you get 7.
Do 38/7 + 7 then divide by 2 and you get 6.2142857 which is already 0.05 units from accuracy.
Now do 38 / 6.2142857 +6.2142857 then divide by two
and you get 6.164614 which is only off by 0.0002 units.
Except use fractions. You get 87/14 from the first iteration.
38 divided by 87/14 + 87/14 then divide by two and
you get (38*14^2 + 87^2) / (2*87*14) from the second iteration
or 15017/2436
This is what the babylonians did and what is still now done by modern calculators.
*THIS* - Yes, I completely agree this is a superior method. In fact, I haven’t used the method shown in the video for years. Since I teach kids ranging from age 9-14, I use a simple version of your above-described method which takes quotient of the sum of the least perfect square less than the sought square and the sought square divided by twice the positive square root of the least perfect square. This method, only one single iteration, almost always produces an approximation accurate to the tenths place-which is about as much precision the kids can do entirely in their heads (none of this is in the Texas math standards so I just teach it as a sidebar and make it a mental math exercise). Even given the limited precision of this method, most parents, administrators, and succeeding teachers are astonished that 95% of the 7th graders refuse to use the calculator because they can compute the root approximation to a sufficient level of precision in their minds to solve the problem faster than it would take them to get up, walk across the room, and fetch the calculator, and moreover, the students feel a much greater sense of accomplishment having done all the math without the aide of a machine.
Edit: The kids and others are quick to notice the approximation becomes increasingly less accurate the closer the sought after square is to the next larger perfect square (or, in the same manner, the closer the sought root’s square is to the largest smaller perfect square e.g. using this method for sqrt(37) and sqrt(80) will be less accurate using this simple version of your method than, say, sqrt(42) because the radicand lies about halfway between the two nearest perfect roots, 36 and 49.
The calculation for approximating √42 using this simple method is to add 36 and 42 == 78 (the numerator) divided by 2 times √36 == 12 (denominator). Kids can mentally do everything above and reduce the resulting fraction of 78/12 to 39/6 and calculate the approximation as ~6.5 within 30 seconds or less. The actual approximation is 6.48074069840786 so if I wrote the problem set asking for precision to the tenths place, the kids are correct 👍.
@@AthenianStranger if all calculators use this, and its only accurate to the 10s place, does that mean our calculators are only accurate to the 10s place?
Finally! I was taught that in 6 or 7th grade back in the late 60s. In college asked my professor for this demonstration, because I forgot. He told me use a calculator, and refused. lol. Thanks! Now I can die!
He might have refused because he didn't know the method.
Idk if you only need an approximation then just use linear approximation.
6+(38-36)/(49-36)=6+2/13approx6.15
I need to review this video several times since I never had square roots in school.
Perhaps my updated and much easier version of finding square roots by hand will be more accessible to you: th-cam.com/video/xIpoxUPfuRY/w-d-xo.htmlsi=Mr3o-MdTBeMC-LvP
I think I saw another method whre you won't end up with numbers you have to calculate getting larger. You should substract some number combination in the process and you end up figuring out only two digits numbers.
real men square root with their fists
Interesting. I learned math in the pre calculator era and was never shown this. Love it.
Me too!
nice to watch..... forgot my high school math, and I remember my first calculator.
So which of the two 6's are always to be doubled? The divisor or the quotient?