@Mr H: On the last example, instead of 100-97, you could use 97-100 (as same as in the first two examples) and this would lead to -3 on the numerator. So the method works the same way for the upper as well as the lower approach.
Check out Newton Raphson method (my 72th birthday today - still remember the essence - made many iterations & approximations since I was young and before I learned more rational approaches - my teachers never understood what I did...).
Alternatively start with the equation of a tangent line (or Taylor expansion) t(x)=f(x0) + f'(x0)*(x-c0) Insert f(x0)=sqrt(x0), and f'(x0)=1/(2*sqrt(x0)) to get t(x)=sqrt(x0) + (x-x0)/(2*sqrt(x0))
@@anotherelvis I like Taylor expansions - such an approach was actually the basis for a short radiation biology publication I and a British scientist wrote in 2002*. It was a relatively simple thing - the key was just to realize that Taylor expansion could be used to make an important point (about fragmentation of irradiated DNA depending on chromosomal sizes and related to the species). * The Fraction of DNA Released on Pulsed-Field Gel Electrophoresis Gels may Differ Significantly between Genomes at Low Levels of Double-Strand Breaks. Radiation Research, Vol. 158(2), pp. 247-249.
Thx! 👍 I had long forgotten how to do this. 😄 Since school, I've mostly just guesstimated the difference between the radicand and its closest perfect squares. 😄
This can be generalised into nth root of (x^n + a)^(1/n) ≈ x + a/(nx^(n-1)) e.g. cube root of 70 = (4^3+6)^(1/3) ≈ 4 + 6/((3)(4²)) = 4.125. The actual value is 4.121 to 3 decimal places
@@homosapien6031 For me this is Calc 1 but our curriculums may be different - it's basically drawing a tangent line at a close known value and using that tangent line to approximate the answer (therefore, linear approximation). I've seen this method before from another channel (maybe it was bprp) but only now do I realize that this method follows the formula used for linear approximation.
@@hywell_ ohh I vaguely remember that thanks. I took AP Calc BC a few years ago in truth. That covers up to Calc 2 so that’s why I said that, but it’s been a few years, so I forgot I learned that
this is the first term of the taylor expansion of sqrt(1+x). Assume we have sqrt(b) if you take the closest square out, lets call it a then the rest will be sqrt( a^2 + (b - a^2)) = sqrt(a^2)*sqrt( 1 + (b-a^2)/a^2). Now we have in in form of the taylor expansion. (basically approximating a function with a polynomial that have the same y value , derivative, second derivative ect. ) The taylor exppansion of sqrt(1+x) is 1 + x/2 - x^2/8 ... substituting the formula derived earlier we have (and only using the first term because the series converges rapidly) 1 + b-a^2/2a^2, which is the formula of the video.
Cool, I haven’t done square roots since school in the early 1970”s.😊 and we didn’t have calculators just paper ,pencil , and slide rule. I didn’t get a calculator til I was in college. Part of me wants to learn advanced math again. Now that I’m older, things make more sense 😂 I’ve subscribed.
Very good and will be occasionally useful. I enjoyed your presentation. What kind of audio system are you using? I found the audio somewhat muffled due to an apparent deficiency in the high frequency range. Therefore, it would have been better if I could have understood what you said better.
I have never seen this method before, but you still have to deal with fractions which you may still have to use a calculator to take care of the fraction portion. I recommend using the binomial expansion method.
Here is a reason to do this even if you have a calculator: if you are, for whatever reason, trying to find the difference between sqrt(64) and sqrt(64.000001). Your calculator may have trouble calculating that, or at least displaying it with enough precision.
Sir, I have a question. In the denominator of decimal part it is 2 multiplied by the given number. Had this been the case of cube root finding, then would we use 3 in place of 2 in the decimal part?
Yes indeed. In a general sense for finding the nth root with a method similar to this, we rewrite this nth root as a power of 1/n. nth root of x = x^(1/n) Then, we use the power rule to take the derivative: d/dx x^p = p*x^(p - 1) Apply for p = 1/n: d/dx nth root of x = 1/n*x^(1/n - 1) So as you can see, this is where the 1/2 comes form for square roots, and where 1/3 would come from for cubic roots. Plug in x = x0, for the reference point. Call this k. k = d/dx x^(1/n) at x = x0: k = 1/n*x0^(1/n - 1) The tangent line L(x) to the original x^(1/n) function, will therefore be: L(x) = k*(x - x0) + x0^(1/n)
It is important to keep in mind - you are not actually finding the square root - you are approximating the square root. The approximation it the worst what the number is between two squares; for example 156 is 12 more than 144 or 13 less than 169, giving you approximations of 12.500 either way; but the actual root is 12.490. There are two methods that you can use to calculate a square root - one is Newton's method (using a method of his from Calculus, though the actual method was known in Babylonian times) and another is called the 'long-division method', so named because it looks kind of like long division. Newton's method is very fast, and pretty easy to do. Guess what the root of your number n is (say r_1), then find the average of r_1 and n/r_1; call this number r_2. Repeat - find the average of r_2 and n/r_2. Continue until you are as close as you need to be. For 150, we can start with 12: average 12 and 150/12: 1/2 * (12 + 150/12) = 1/2 * (144/12 + 150/12) = 1/2 * (294/12) = 147/12 = 12.250 But if we do that again we get: 43209/3527 = 12.24744898, the actual value being 12.24744871 -- off in only the last two decimal places! You need to work it out as fractions - but you double the number of accurate decimal places with each time you do it. The 'long division' method is a bit more difficult to show in a comment like this, but it has the advantage of stopping if you have a perfect square, and everything stays as a decimal. It adds one decimal point with each row, basically, but - each additional decimal point takes a little more work.
@@paulstudier5706 Using a slide rule what you are actually doing though is taking the log of the number, and dividing it in half. Then raising the base to the power of the answer. For example, log to the base 10 of 156 is 2.19312 and 10 to the power of half of that is 12.49. 12.49*12.49=156.0001 Close enough
It can also be seen as use of the Newton's method for x²-65=0. Just the signs are slightly different. That also explains why the constant is 2. In the Newton's method it would be 8 - (64-65) / (2*8)
This comes from finding the derivative (i.e. slope of the tangent line) of y=sqrt(x), and then using it to construct a tangent line called linearization, to approximate it in the neighborhood of a known square root. For a general power function, y=x^p, the derivative is: d/dx x^p = p*x^(p - 1) For square roots, p=-1/2. For roots in general, the power p = 1/n. Plug in p=1/n and get: d/dx x^(1/n) = (1/n)*x^(1/n - 1) Evaluate at your reference point, x0, and call this slope k: k = 1/n*x0^(1/n - 1) Build your linear approximation L(x) from k and the reference point: L(x) = k*(x - x0) + x0^(1/n) For n=2 for square roots: k = 1/2*x0^(1/2 - 1) = 1/(2*sqrt(x0)) L(x) = 1/(2*sqrt(x0)) * (x - x0) + sqrt(x0) For n=3 for cube roots: k = 1/3*x0^(1/3 - 1) = 1/(3*cbrt(x0)^2) L(x) = 1/(3*cbrt(x0)^2) * (x - x0) + cbrt(x0)
@@gregc.mariano9226 sorry for hurting your feelings but I keep what I said. It is not about mind or superminds, it about paper + pencil and any method you learned in school.. if you don’t know fractions why are you trying to solve square roots without calculator 😅
@@gregc.mariano9226 Yes i kind of agree with what you saying, but if you use calculator to calculate a decimal of a fraction then shouldnt you just use it to calculate sqrt as well? No offence but I just questioned this
This is the first term of the Taylor series, which is called linear approximation. You construct a tangent line at the reference point, and use it to approximate a square root in the same neighborhood.
I think it’s a the first term of the expansion of its Taylor series. What does this actually mean? Well, each function, like sqrt(x) can be represented as an infinite series. Each term of the series is a fraction, and each successive term of the fraction is smaller and smaller in value. So if you just use the first term, you’ll already be close enough to have a good approximation. I don’t actually know how that works (still haven’t had Taylor series, but did have the series of e^x, sin(x) and cos(x), and another commenter might’ve put the thought in my mind. You can look up “Taylor Series” on Wikipedia and try to change the language to “simple English” for a better understanding. Hope this helps!
In sqrt(x), he made a tangent line on the point where x = 64, then he took the derivative of sqrt(x), which is 1/2sqrt(x), then he plugged it in, 1/2sqrt(64), which means the tangent line is x/16, and since 65-64 = 1, you do 1/16 * 1, and added to 8, and you get 8 1/16. This also means that the larger the number is, the better it approximates it.
Incrivel. Calculei como seria o 81 eu posso fazer isso? Ou é só o número mais proximo? Incredible. I calculated what 81 would be like. Can I do this? Or is it just the closest number?
Nice trick**, but (@MrH) you better should have called it _estimate_ instead of _finding_ the sqrt, don't you think? 🙂👻 ** more a clever idea using the derivative of the sqrt-function for a pretty good estimate rather than a 'trick'.
First: Use the nearest sqare, which is 36. => sqrt(35) ~= 6 - 1/12 = 5.91666... Second: For b = 5 and d = 10, use my method above with d/2b = 1. d²/8b³ = 1^3 / 10 = 0.1. Subtract this from 6 and get 5.9 (even with b = 5). Exact: 5.916079...
Well… In this approximation technique, you’re supposed to start with the CLOSEST perfect square to the number you’re trying to find the square root of. The closest perfect square to 36 is… 36 (= 6^2) with a difference of 0. So, this technique give approximation: 6 + 0/(2*6) = 6 + 0 = 6, which is as close to the square root of 36 as you could get!! 😂
When you see fractions with denominators that easily divide into a power of ten, get an equivalent fraction by multiplying top & bottom by the number that will get that power of 10 in the bottom. In fraction 3/20, 20 goes into 100 five times. So, multiplying top & bottom each by 5 gives equivalent fraction 15/100. Multiplying or dividing by powers of 10 is easy: move decimal as many places as that power of ten has zeroes, in the direction that makes sense. Here 15 is divided by 100, which has two zeroes. So, we move the decimal of 15. two places TO THE LEFT (b/c dividing by 100 makes a number smaller, and moving decimal to the left makes a number smaller…). Without all the talking: 3/20 = = (3*5)/(20*5) = 15/100 = 0.15 Done!
8 x 8 = 64 8.1 x 8.1 = 64.8 + 0.81 = 65.61 8 < ✓65 < 8.1 8.05 x 8.05 = 64.4 + 0.4025 = 64.8025 8.05 < ✓65 < 8.1 continue until you have obtained the number of digits to your satisfaction
For the square root of a non-perfect square number you still need a calculator like the 1/16. Others might have a hard time doing this without a calculator. If you have memorized the all the decimal number equivalent of fractions then so be it, otherwise, you have to do the manual division.
Think of a real square. X is the initial guess. Y is what must be added. So the area is (x+y)^2=x^2+2*x*y+y^2. What is added is the 2*x*y so after one iteration the estimate is short by y^2. Another iteration will be closer yet but the new estimation is always short by the current value of y^2.
@Mr H: On the last example, instead of 100-97, you could use 97-100 (as same as in the first two examples) and this would lead to -3 on the numerator. So the method works the same way for the upper as well as the lower approach.
Thats very helpful,thank you!!
I'd like to see a geometric explanation of why this approximation works.
Check out Newton Raphson method (my 72th birthday today - still remember the essence - made many iterations & approximations since I was young and before I learned more rational approaches - my teachers never understood what I did...).
Alternatively start with the equation of a tangent line (or Taylor expansion)
t(x)=f(x0) + f'(x0)*(x-c0)
Insert f(x0)=sqrt(x0), and f'(x0)=1/(2*sqrt(x0)) to get
t(x)=sqrt(x0) + (x-x0)/(2*sqrt(x0))
@@anotherelvis I like Taylor expansions - such an approach was actually the basis for a short radiation biology publication I and a British scientist wrote in 2002*. It was a relatively simple thing - the key was just to realize that Taylor expansion could be used to make an important point (about fragmentation of irradiated DNA depending on chromosomal sizes and related to the species).
* The Fraction of DNA Released on Pulsed-Field Gel Electrophoresis Gels may Differ Significantly between Genomes at Low Levels of Double-Strand Breaks. Radiation Research, Vol. 158(2), pp. 247-249.
@@bjorncedervall5291 cool!
Thx! 👍 I had long forgotten how to do this. 😄
Since school, I've mostly just guesstimated the difference between the radicand and its closest perfect squares. 😄
Awesome teaching, as always with Mr H.
This can be generalised into nth root of (x^n + a)^(1/n) ≈ x + a/(nx^(n-1))
e.g. cube root of 70 = (4^3+6)^(1/3) ≈ 4 + 6/((3)(4²)) = 4.125. The actual value is 4.121 to 3 decimal places
You are an awesome teacher👍🏻🇺🇸
This is really cool!
Just learned linear approximation recently so it finally makes a bit more sense to me now why this method works.
Just curious. Is that linear algebra or…?
Asking cuz I’ve taken up to Calc 2 so far and haven’t really gotten there ywt
@@homosapien6031 For me this is Calc 1 but our curriculums may be different - it's basically drawing a tangent line at a close known value and using that tangent line to approximate the answer (therefore, linear approximation). I've seen this method before from another channel (maybe it was bprp) but only now do I realize that this method follows the formula used for linear approximation.
@@hywell_ ohh I vaguely remember that thanks. I took AP Calc BC a few years ago in truth. That covers up to Calc 2 so that’s why I said that, but it’s been a few years, so I forgot I learned that
this is the first term of the taylor expansion of sqrt(1+x). Assume we have sqrt(b) if you take the closest square out, lets call it a then the rest will be sqrt( a^2 + (b - a^2)) = sqrt(a^2)*sqrt( 1 + (b-a^2)/a^2). Now we have in in form of the taylor expansion. (basically approximating a function with a polynomial that have the same y value , derivative, second derivative ect. ) The taylor exppansion of sqrt(1+x) is 1 + x/2 - x^2/8 ... substituting the formula derived earlier we have (and only using the first term because the series converges rapidly) 1 + b-a^2/2a^2, which is the formula of the video.
Excellent, thank you.
Thank you sir
woah, that was not discussed in my math lessons. but definitely should be in curriculum !
very nice
Fantastic - another great method I can teach my students.
Cool, I haven’t done square roots since school in the early 1970”s.😊 and we didn’t have calculators just paper ,pencil , and slide rule. I didn’t get a calculator til I was in college. Part of me wants to learn advanced math again. Now that I’m older, things make more sense 😂 I’ve subscribed.
Very good and will be occasionally useful. I enjoyed your presentation. What kind of audio system are you using? I found the audio somewhat muffled due to an apparent deficiency in the high frequency range. Therefore, it would have been better if I could have understood what you said better.
Thank you, sir.
Thank you so much Sir ❤
chad math teacher❤
Thank you Sir.
awesome
Wow thank you
Thank you so much 😭🙏
I have never seen this method before, but you still have to deal with fractions which you may still have to use a calculator to take care of the fraction portion. I recommend using the binomial expansion method.
Great apprauch thanks v much
He is using the derivative of X^0.5 to find the fractional part added to the perfect square. Easy!
Thank you
Here is a reason to do this even if you have a calculator: if you are, for whatever reason, trying to find the difference between sqrt(64) and sqrt(64.000001). Your calculator may have trouble calculating that, or at least displaying it with enough precision.
I'm not smart, but this made me feel like I have the potential to be smart. 🙏
Sir, I have a question.
In the denominator of decimal part it is 2 multiplied by the given number.
Had this been the case of cube root finding, then would we use 3 in place of 2 in the decimal part?
Yes indeed.
In a general sense for finding the nth root with a method similar to this, we rewrite this nth root as a power of 1/n.
nth root of x = x^(1/n)
Then, we use the power rule to take the derivative:
d/dx x^p = p*x^(p - 1)
Apply for p = 1/n:
d/dx nth root of x = 1/n*x^(1/n - 1)
So as you can see, this is where the 1/2 comes form for square roots, and where 1/3 would come from for cubic roots.
Plug in x = x0, for the reference point. Call this k.
k = d/dx x^(1/n) at x = x0:
k = 1/n*x0^(1/n - 1)
The tangent line L(x) to the original x^(1/n) function, will therefore be:
L(x) = k*(x - x0) + x0^(1/n)
It is important to keep in mind - you are not actually finding the square root - you are approximating the square root. The approximation it the worst what the number is between two squares; for example 156 is 12 more than 144 or 13 less than 169, giving you approximations of 12.500 either way; but the actual root is 12.490.
There are two methods that you can use to calculate a square root - one is Newton's method (using a method of his from Calculus, though the actual method was known in Babylonian times) and another is called the 'long-division method', so named because it looks kind of like long division.
Newton's method is very fast, and pretty easy to do. Guess what the root of your number n is (say r_1), then find the average of r_1 and n/r_1; call this number r_2. Repeat - find the average of r_2 and n/r_2. Continue until you are as close as you need to be.
For 150, we can start with 12: average 12 and 150/12:
1/2 * (12 + 150/12)
= 1/2 * (144/12 + 150/12)
= 1/2 * (294/12)
= 147/12 = 12.250
But if we do that again we get:
43209/3527 = 12.24744898, the actual value being 12.24744871 -- off in only the last two decimal places!
You need to work it out as fractions - but you double the number of accurate decimal places with each time you do it.
The 'long division' method is a bit more difficult to show in a comment like this, but it has the advantage of stopping if you have a perfect square, and everything stays as a decimal. It adds one decimal point with each row, basically, but - each additional decimal point takes a little more work.
Use a slide rule to get the square root to 2 or 3 decimal places, then apply this method. This saves an iteration.
@@paulstudier5706 Assuming you have a slide rule (which _I_ do, but not everyone does...)
Hahahahaha. You saying
"you are not actually finding the square root - you are approximating the square root"
is so funny. Hahahahahahahahahaha.
@@paulstudier5706 Using a slide rule what you are actually doing though is taking the log of the number, and dividing it in half. Then raising the base to the power of the answer. For example, log to the base 10 of 156 is 2.19312 and 10 to the power of half of that is 12.49. 12.49*12.49=156.0001 Close enough
Where does this come from? The theory please.
This method is derived from the approximation of the first two terms of the binomial expansion (a+x)½ where a is the required perfect square.
It can also be seen as use of the Newton's method for x²-65=0. Just the signs are slightly different. That also explains why the constant is 2. In the Newton's method it would be 8 - (64-65) / (2*8)
Square numbers until you get close.
Why is the constant always 2 in the denominator? And would this metbod work with cube root?
Because
(√x)' = 1/(2√x)
This comes from finding the derivative (i.e. slope of the tangent line) of y=sqrt(x), and then using it to construct a tangent line called linearization, to approximate it in the neighborhood of a known square root.
For a general power function, y=x^p, the derivative is:
d/dx x^p = p*x^(p - 1)
For square roots, p=-1/2. For roots in general, the power p = 1/n. Plug in p=1/n and get:
d/dx x^(1/n) = (1/n)*x^(1/n - 1)
Evaluate at your reference point, x0, and call this slope k:
k = 1/n*x0^(1/n - 1)
Build your linear approximation L(x) from k and the reference point:
L(x) = k*(x - x0) + x0^(1/n)
For n=2 for square roots:
k = 1/2*x0^(1/2 - 1) = 1/(2*sqrt(x0))
L(x) = 1/(2*sqrt(x0)) * (x - x0) + sqrt(x0)
For n=3 for cube roots:
k = 1/3*x0^(1/3 - 1) = 1/(3*cbrt(x0)^2)
L(x) = 1/(3*cbrt(x0)^2) * (x - x0) + cbrt(x0)
Shame on you for those who say “still need calculator for fraction” 😮
Shame on you too. Others' minds are not the same as yours (coconut). Pls don't say those words as it's an insult to others.
@@gregc.mariano9226 sorry for hurting your feelings but I keep what I said. It is not about mind or superminds, it about paper + pencil and any method you learned in school.. if you don’t know fractions why are you trying to solve square roots without calculator 😅
@@gregc.mariano9226 Yes i kind of agree with what you saying, but if you use calculator to calculate a decimal of a fraction then shouldnt you just use it to calculate sqrt as well? No offence but I just questioned this
Hahaha
Lol
This is not finding the squareroot, but doing an approximation.
Since it's an approximate answer 8 is acceptable. There isn't anything saying where to round to.
It’s tricky but you don’t really give the mathematical explanation. Maybe the Taylor expansion !
This is the first term of the Taylor series, which is called linear approximation. You construct a tangent line at the reference point, and use it to approximate a square root in the same neighborhood.
Subtle Way...thanks
So, why does this actually work? I understand how but not why…
I think it’s a the first term of the expansion of its Taylor series.
What does this actually mean?
Well, each function, like sqrt(x) can be represented as an infinite series. Each term of the series is a fraction, and each successive term of the fraction is smaller and smaller in value. So if you just use the first term, you’ll already be close enough to have a good approximation.
I don’t actually know how that works (still haven’t had Taylor series, but did have the series of e^x, sin(x) and cos(x), and another commenter might’ve put the thought in my mind. You can look up “Taylor Series” on Wikipedia and try to change the language to “simple English” for a better understanding. Hope this helps!
In sqrt(x), he made a tangent line on the point where x = 64, then he took the derivative of sqrt(x), which is 1/2sqrt(x), then he plugged it in, 1/2sqrt(64), which means the tangent line is x/16, and since 65-64 = 1, you do 1/16 * 1, and added to 8, and you get 8 1/16.
This also means that the larger the number is, the better it approximates it.
Incrivel. Calculei como seria o 81 eu posso fazer isso? Ou é só o número mais proximo?
Incredible. I calculated what 81 would be like. Can I do this? Or is it just the closest number?
This is just Newton's method for successive approximaion of a square root.
Nice trick**, but (@MrH) you better should have called it _estimate_ instead of _finding_ the sqrt, don't you think?
🙂👻
** more a clever idea using the derivative of the sqrt-function for a pretty good estimate rather than a 'trick'.
what happened to Route 66?
That's a good one!
√35=√(25+10)
=√25+10/(2×5)
. =5+1
. =6
But 6²=36🤔
This is an approximation. I am pretty sure since root35 is 5.9, it vecomes 6
You have violated the rule of nearest perfect square. the nearest perfect square to 35 is 36
@@superacademy247 he probably knows, fool.
First: Use the nearest sqare, which is 36. => sqrt(35) ~= 6 - 1/12 = 5.91666...
Second: For b = 5 and d = 10, use my method above with d/2b = 1.
d²/8b³ = 1^3 / 10 = 0.1. Subtract this from 6 and get 5.9 (even with b = 5). Exact: 5.916079...
what about if the number is less than one, say sq root of 3/10
You'd take square root of 3 and 10 separately.
Or:
3/10 = 0.3 = 30/100
So:
take square root of 30
and then
divide by 10 (sqrt of 100)
My lab math: 0/1 points
I beg to mention this process has limitation.
As for example
According to ur process
√36=5 + 11/5*2=6.1
It is higher than actual so root of 36
Well… In this approximation technique, you’re supposed to start with the CLOSEST perfect square to the number you’re trying to find the square root of. The closest perfect square to 36 is… 36 (= 6^2) with a difference of 0. So, this technique give approximation:
6 + 0/(2*6) = 6 + 0 = 6, which is as close to the square root of 36 as you could get!! 😂
@@timeonly1401 I agree. Thanks.
This method was discovered by Emmy Noether when she was at school.
nope, issac newton discovered the generalized binomial theorem a few centuries earlier, this is just the first term
@@t.b.4923 this method exists since ancient times, fool.
In the last example many people will stuck in that fraction 3/20..is there any simple method to convert this to decimal?
You can change 3/20 into (300/20)*(1/100) which can simplified into (30/2)*(1/100) and that will be 15/100 which will be 0.15
Recongize that if it were 3/10, that it would be 0.3. So cut 0.3 in half, to get 0.15.
When you see fractions with denominators that easily divide into a power of ten, get an equivalent fraction by multiplying top & bottom by the number that will get that power of 10 in the bottom.
In fraction 3/20, 20 goes into 100 five times. So, multiplying top & bottom each by 5 gives equivalent fraction 15/100. Multiplying or dividing by powers of 10 is easy: move decimal as many places as that power of ten has zeroes, in the direction that makes sense. Here 15 is divided by 100, which has two zeroes. So, we move the decimal of 15. two places TO THE LEFT (b/c dividing by 100 makes a number smaller, and moving decimal to the left makes a number smaller…).
Without all the talking:
3/20 =
= (3*5)/(20*5)
= 15/100
= 0.15
Done!
8 x 8 = 64
8.1 x 8.1 = 64.8 + 0.81 = 65.61
8 < ✓65 < 8.1
8.05 x 8.05 = 64.4 + 0.4025 = 64.8025
8.05 < ✓65 < 8.1
continue until you have obtained the number of digits to your satisfaction
*uses calculator to calculate the fractions
No need for calculator for that either.
@@mrhtutoring right 😅
That's approximating, not "finding" the roots.
Still need calculator for the Fraction
But still it is a crazy method🎉
For the square root of a non-perfect square number you still need a calculator like the 1/16. Others might have a hard time doing this without a calculator. If you have memorized the all the decimal number equivalent of fractions then so be it, otherwise, you have to do the manual division.
Think of a real square. X is the initial guess. Y is what must be added. So the area is (x+y)^2=x^2+2*x*y+y^2. What is added is the 2*x*y so after one iteration the estimate is short by y^2. Another iteration will be closer yet but the new estimation is always short by the current value of y^2.
Another iteration will increase the accuracy.
Make on how to divide ➗ quick
unfortunately it's good old "long division"
Thank you