Video

I favore the method of Heron. Take an initial guess, let s say 300. Compute (300+84 765.2347/300)/2 Repeat the process once and you already obtain the same precision than in the video... even one more digit in fact.

This is a well known method for calculating the square root and whilst the steps were clearly explained, the alignment of the numbers was poor. It would help of the answer digits were spaced out to cover the corresponding pairs below. It would then be obvious where the decimal point belongs. Similarly, the numbers on the left belong opposite the numbers to which they correspond to avoid confusion.

It doesn’t give you an exact answer when you try it with 654

I keep getting stuck trying to find the square root of 41 because the number on the side is greater than the number on the bottom

what kind of a shit crank number is 64527345.856378, fam im tryna do 80

is the first red number always 2?

i feel like. this is one of those videos where it clicks on the 2nd-3rd view

Ug ihc😢😮😅🎉❤😅😊😮😮😮😮😮😮

Why don't they teach this anymore??? Most people now believe that squareroots are impossible to take by hand. It's weird how common knowledge that isn't included in school gets phased out...

Thanks a lot sir really helped me.

What if the thousands digit is a 1?

Easy proof more like impossible proof

It breaks when the number starts with 1

Here's how it works. Let's say that we have a guess on the square root "r", and we know it's too low, but by an unknown amount. Let's call that unknown amount "a". So we can say that (r+a)^2 will give us the number we want to take the square root of (the radicand). We can also say that (r+a)^2 - r^2 = 2ra+a^2, or a(2r+a). So what this algorithm is doing is coming up with informed estimates on "a" and using them to refine our "r". It's an iterative process, where on each iteration we tack another digit on to "r", while the "a" is covering smaller and smaller inexactitudes. You can see us trying to make informed estimates on "a" at 3:25-4:00, where we decide that a(2r+a) works best if our "a" is a 9. It's still not a perfect guess, but it's good enough to give us one more digit on "r". Then on the next pass, we're working with a smaller "a".

This is easy???????????

as a math lover this hurt me, it feels illegal to make math so complicated

Wow I'm so glad I found this video. I never learned this technique in school this is way easier and this vid is 7yrs old already. It took me too long to discover this. Thank you so much! Math is indeed fun!

who was this found by?

I learned another way with zeros. Its hard to go against my learning. Im old. I will give this a chance. Thanks.

答え　6^ー2

This method should be useful by square roots without calculator!!!! Here I go, Math!!! Cuz here comes the MATH HUMAN CALCULATOR!

That is what we were taught.

You realize that the "numbers in black" 4/58/582... Is just doubling what you find as the answer for that step? 2=2x2=4 29=29x2=58 291=291x2=582 2911=2911x2=5822 29114=29114x2=58228

This is the method I learned in 6th grade a looonngggg time ago. If you think of the dividend starting with zero then the step where you put the red two is simply following the algorithm that is used in subsequent steps.

this is what we do in school

come back chief

Fascinating! Can someone mention which mathematician came up with this method and if there's a paper with the derivation or sth?

Okay, so this is a _method_. Clear enough. But I’m puzzled about WHY anyone would want to take a square root of a number such as this, other than as an arbitrary exercise. No one has.explained to me the REASON for deriving a square root. Most arithmetic or algebraic problems involve discovering a value not currently known. But deriving a number that can be multiplied by itself to produce a given number seems sterile.

I can’t do is with sqer 2 It give me 1.6156

291.1446971868 exact answer Just do calculator

what if there is an odd number of decimals? when do i add the decimal to the square root

It's the only method people should use . It reminds me of how people struggled to balance equations

Already done it in school 👍🏻 Nicely explained.

Hi! Great explanation. How would this work for single-digit numbers? Thanks!

Nice video! Thank you.

hl

I am 13 and i learnt this at my school a week ago

Good

It appears that the number of significant digits of the root must be the same or more than the starting number for it to be accurate near the ones digit.

Ima better with derivative approximation after all I don't like long division

Never seen this method before!

Whatever. Root is a split. What are the directions. The mean.

I've reached a point where no matter for what number I would multiply it, the number on the left would always be bigger than the number on the bottom, what do I do? (The situation was: 582 on the left and 5123 on the bottom)

I think it's called "long division method"..

So I have a question so I decided to try this algorithm with my own number of 1,640,299.8978 and it’s going good until I get to the third set of 2 numbers, 02 which if I put at the end of 20 gives me 2002, the number I got from the last one was 24 so with some math I was able to find out that the largest number I could make without going over was 248 times 8, which gives me 1984 and if I subtract it from 2002 I get 18, now I bring the next set of numbers down and I get 1899, strangely smaller then the last, and then if I add 1 to the number I get 249, which if I add any number to the end I will have a number bigger then 1899, I was wondering if someone could help me with this

Best process for square roots: a^2 + 2ab + b^2 a = your currently completed root, b = your next digit In practice, multiply your a by 20 , 20a. That is a partial divisor into your current remainder. With that determine what the largest next digit, b, could be. Add that b to your 20a figure, multiply that by your b. Is that less than your remainder but as large as it could be? If so, subtract from your remainder and continue with your next iteration Very simple and easy to remember __________________________________________________________ For cube roots: x = a^3 + 3ab^2 + 3a^2b + b^3. In practice you do this: In your workspace, Square the current (a) multiply it by 300. 300a^2 Beside that multiply your (a) by 30. 30a Determine your next (trial) digit (b) by mentally using that 300a^2 figure as the main division into your current remainder. Eventually you will become very proficient at getting it right the first time Add your new (b) to the 30a figure, multiply that by the (b), add that to the 300a^2 figure, and multiply by (b) again. Put beneath the current remainder and subtract Do this over and over until you reach your desired accuracy Way more complicated than the sqrt algorithm, but it is the best way Notwithstanding what you see in the vid, these are actually overall, the best methods

Any? What is the square root of -1.23??

This looks like a good clean method, but it is by rote. You arent going to remember it well, because you dont really understand why you are doing what you are doing' The old method is the best. In each iteration just multiply the currently completed root by 20. That is your divisor. Add you next digit to it, multiply that by that next digit. Done! Subtract that from the current remainder in the bottom of your "long division" . Proceed with the next iteration There is really little to remember in the old tried and true method

if its a whole number and you need to keep going do you put a decimal then the 0s?

In Iran, we were not allowed to use calculators, and I believe it is still the case there. You might laugh, but we had to learn squares and square roots of 1-20, sin, cos, tan and cotans of round angles etc etc by heart. We were taught how to take square roots when we were 12 or 13, and a version of this was the method they taught us. I remember I found it a bit complicated at the time, and had to revisit it for use at high school. The version we were taught was that to get the black numbers on the left, you double the numbers above the bar. Also, to get the red numbers, you bring down only one digit first and then you divide that number by the number on the left, i.e. 44/4=9; 66/58=1; 842/582=1; 26024/5822=4. Then you bring the next number down.