How to calculate square roots in your head... no aspirin required.
Square roots are very useful and at times one wishes you could calculate them easily without a calculator. Well, while I have not found a way to do this for all values, I came-up recently with an easy way to do this for values below 100. This method assumes you know multiplication tables by heart (do they still teach them in school these days? I wonder...). By this I mean that you can calculate 2x2=4 and 3x3=9 and 4x4=16 ... up to 9x9 = 81 without any effort. It is also assumed you only need an approximation of the exact value... say within 5% or so. This is ok of most practical cases. If you need more precision than this, you probably shouldn't be doing this in your head... (time to reach for a calculator).
If you can compute squares as described above by heart (2x2 and 3x3 and 4x4 ... up to 9x9) then you already know the first digit of any square root below 100. Let's say for example you want to calculate √40. Well, since 6x6 = 6^2 = 36 < 40, you know √40 must be more than 6. On the other hand, 7x7 = 7^2 = 49 > 40 so √40 must be less than 7.
So, by looking at the square values immediately below and above the desired value, you identify the first digit for the square root. We can safely assume √40 = 6.(xxx). The trick now is to find the second digit and that involves some (rather simple) math.
Let's say √x = (a + b)
Where a is the integer part (the one we already calculated) and b is the fractional part (the one we need now to approximate). Expanding the equation we get:
x = (a + b)^2 = a^2 + b^2 + 2ab
Since b is the fractional part (b<1) then we can recognize that b^2 will be a very small number relative to the other two terms above and we can drop it to get the following approximation:
x = a^2 + 2ab = a^2 + (2b).a
The formula above is the tool we need to get a better approximation of the square root. Now, in practice we don't need a very exact approximation and we can live with three cases for the "fractional" part of the square root (b). Let's consider the following three cases:
b = 0.25 => (2b).a = 0.5a (in other words, add half of a to the result)
b = 0.5 => (2b).a = a (in other words, add a to the result)
b = 0.75 => (2b).a = 1.5a (in other words, add one and a half times a to the result)
So to summarize, once we determine the first digit a, we choose a fractional part b of either 0.25, 0.5 or 0.75, and add half of a, a, or one and half times a respectively to a^2 to see which comes closer to x. The number a+b that comes closer to x using the formula above is your approximation.
I know this might sound complicated now, but things will be clearer with a couple examples. It is really quite simple...
With √40 (so x = 40) we already determined that a = 6 since 6x6=36 is the square that comes closer without exceeding 40. Now if we choose b = 0.25, we get per the equations above:
b = 0.25 => we add half of a = 3
and 36 + 3 = 39
This is already a pretty close approximation to 40 so one could approximate √40 = a+b = 6.25
In reality, √40 = 6.3245... an error of only 1.2 %
First we find that the first digit a = 8 since 8x8 =64 (and 9x9 = 81 which is already too much).
Now we choose b = 0.5 so we have to add a according to the formulas above:
64 + a = 64 + 8 = 72
And we stop here to claim our approximation that √72 = a + b = 8.5
In reality, √72 = 8.4852... an error of only 0.17 %
The first digit a must be 6 since 6x6 =36 < 40
Since just adding a ( in other words b =0.5) doesn't get us close enough to 45 (36 + 6 = 42) we try setting b=0.75 which means we must add 1.5a = 1.5x6 = 9. Since 36 +9 = 45 we can say that b = 0.75 and set our approximation to:
√45 = a+b =6.75
In reality, √45 = 6.7082... an error of only 0.62 %
Once you practice, this becomes quite fast and as you can tell from these examples, the approximations are quite good in percentage terms.
Comments, questions, suggestions? You can reach me at: contact (at sign) paulorenato (dot) com