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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.


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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Whether one place is enough depends on the problem. Often it is, but not always. I don't know how to give a rule. – Ross Millikan Aug 17 '13 at 14:50
You can find a lot of useful information on Wikipedia: Methods of computing square roots – Zaz Mar 3 '15 at 15:17
up vote 11 down vote accepted

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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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 $\sqrt{81}+\frac{18}{2\sqrt{81}} = 10$. – Guntram Blohm Aug 12 '15 at 22:27

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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