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I have 2 questions here.

  • What is the most effective and easy way of calculating square roots in your head to an accuracy of 1 decimal point? This would need to work with at least two digit, non-perfect squares and would have to be doable mentally.

    • How would the method work?
  • Is one decimal place accurate enough for all intents and purposes that you may come across in average math? When would you need more precision?

I am curious because often times in school, I would spend time with paper and pencil, working out approximations for non-perfect squares, it would help if I could get an approximation quicker.

Thanks!

Edit: Please ensure that the method is easy to understand, and to do mentally for a student in grades 8 and above. I don't want to be just memorizing formulas without understanding how they work, I think it would be beneficial for anyone browsing this question.

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  • $\begingroup$ Whether one place is enough depends on the problem. Often it is, but not always. I don't know how to give a rule. $\endgroup$ – Ross Millikan Aug 17 '13 at 14:50
  • $\begingroup$ You can find a lot of useful information on Wikipedia: Methods of computing square roots $\endgroup$ – Zaz Mar 3 '15 at 15:17
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I'd just do good old Taylor series expansion up to linear term. $$ f(x+\Delta x) \approx f(x) + f'(x) \Delta x $$ so, in case of square roots $$ \sqrt{x + \Delta x} \approx \sqrt x + \frac {\Delta x}{2\sqrt x} $$ where $x$ - is the closest perfect square. Obviously, error might be huge if $\Delta x$ is big.

For example, $$ \sqrt{66} = \sqrt{64 + 2} \approx \sqrt{64} + \frac 2{2 \sqrt{64}} = 8 + \frac 18 = 8.125 $$ whereas $\sqrt{66} \approx 8.12403840463596 \ldots$

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  • 1
    $\begingroup$ And the same with subtracting if your number is just a bit lower than a square - $\sqrt{100}-\frac{1}{2\sqrt{100}} = 9.95$ is way better for $\sqrt{99} \approx 9.949874...$ than $\sqrt{81}+\frac{18}{2\sqrt{81}} = 10$. $\endgroup$ – Guntram Blohm supports Monica Aug 12 '15 at 22:27
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You should know all the perfect squares up to $100$, and the fact that $(n+\frac 12)^2=n(n+1)+\frac 14$. Then your best friend is $\sqrt{1+x}\approx 1+\frac x2$ for $x\ll 1$ Say you want the square root of $72$. You could say $72=8*9$, so the square root of $72.25$ is $8.5$ The odd quarter doesn't matter. If you want the square root of $68$, you have that $68=64(1+\frac 1{16})$, so $\sqrt{68}\approx 8(1+\frac 1{32})=8.25$ As the approximation is a little high, this rounds down to $8.2$, but it is close.

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There is a method i used to find about 4 decimals which works pretty easy, as soon as you understand it... and you need to now the basic squares from 1..10

You make use of sqr(n+k) = sqr(n) + 2*k*n + sqr(k) this is simply pythagoras, now we use that for step by step approximation.

you start with the number you want the root of, lets say 10; the closest lower square is 9 (sqr(n) in the formula), so the root must be 3.xxx (n=3 ; k=0.xxx)

now you subtract the low square from the starting square, 10 - 9 = 1 because of 2*k*n you divide 100times the difference by 20*n and round it down to integer, which is your next decimal (most probably..) that would be (100*1)/(20*3)~1

since you know the approximate size i always leave commas away by multiplying by 10, resp. 100

this makes the new n of 31 which makes sqr(30) + 2*30 + sqr(1) = 961

the following illustration is the easiest way for me to understand:

10 ~ 3
-9 00 + 60*1 + sqr(1) = 961
1 00 : 60 ~ 1

1000 ~ 31
-961 00 + 620*6 + sqr(6) = 96100 + 3720 + 36 = 99856
39 00 : 620 ~ 6

100000 ~ 316
- 99856 00 + 6320*2 + sqr(2) = 9998244
144 00 : 6320 ~ 2

in case any of those numbers to the far right go above 1000, you just have to subtract 1 from the multiplier decimal (here 1 / 6 / 2) this is for the fact that for finding the digit itself you ignore its own square that is added in the end, thus it happens that it's one too high.

i did it in this order:

1000 ~ 31. // 100times starting number and low root n from before
- 961. // subtract sqr(n)
39 00 : 620 ~ 6 // divide 100*difference by 20*n, this is your new digit

1000 ~ 31
- 961 00 + 620*6 + 36 // add up 100*sqr(n) + 20*n*quotient + sqr(quotient), this is your new sqr(n)
39 00 : 620 ~ 6

--> new n=316, sqr(n)=99856

This way you could go on infinitely and find digit after digit, but if you do it mentally you get stuck after 3 or 4 digits, which is ok since you asked for only 1 in the first place :)

im really curious if anyone will understand this xD the idea came to me once in a boring bus drive

greetz
Thomas from Switzerland

  • edit

I'm sorry I was in the middle of calculations while writing this It's not pythagoras (embarrassing), it's just the binomial formula where "a" is the 'low root' with cut off decimals and "b" is the upcoming decimal

your goal is hitting the number you want the root of (10), so a^2 + 2ab + b^2 ~ 10, and what you know is a = 3 subtracting a^2=9 gives 2ab + b^2 = 1 since b is relatively small you ignore it for now some rearranging and expanding makes b = 1/2a = 0.167

now you cut off all but one decimal and multiply by 10 to get the next decimal (0.167 --> 0.1 --> 1) a^2 + 2ab + b^2 = 9 + 0.6 + 0.01 = 9.61 = 3.1^2 adding up "a" and "b" (or 10times) is 31, the new "a"

this can be repeatet with a=31, a=316, and so on

what i think is the advantage of this method, is that once you get used to doing it, you don't need to keep the number you want the root of or any of those squares in between the steps, you only have to know the difference and the root, here that would be:
low root - square difference to 10
3 - 1
31 - 39
316 - 144

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