Math - Square rooting...

[Jack Bond]

Yeah... I learned how to do that, but I've only done it maybe two or three times and the answer never came out right... Anyone have any advice or help?

[Icarus]

Without a calculator?

Make a guess. Check it, and try again.

Say, the sqrt of 12. Well, 3^2 = 9, and 4^2 = 16, so it's about 3.x with x probably less than 5. 3.3^2 = 10.89, so try 3.4.

3.4*3.4 =1.36+10.2=11.56.

That's pretty close, so try 3.5^2. That is 1.75+10.5=12.25, so 3.5 is just too big.

From there, you guess another decimal place, say 3.44, and see where that leads you.

[Dante]

Actually, there are quite a few things one can do with square roots to help with them… In fact you can do an incredible amount with square roots without the aid of a calculator. In fact, on the physics GRE we are oft requested to find nasty square roots without the aid of a calculator, so here are a few tricks I’ve devised.

1. Always remember that you can multiply up any terms that are to the same power,
EX 1.

sqrt(2)*sqrt(10)=sqrt(20)=1.41*3.16=4.46

Using this, you can often find the square roots of higher order numbers by factoring them into smaller numbers you do know, for instance, what is the square root of 16? You should almost instantly answer 4, but you could simply things by changing sqrt(16) to sqrt(4)*sqrt(4)=2*2=4, same as sqrt(16)=4. Generally though it is rather useful for questions such as the following, what is the sqrt(10000), well, sqrt(10000)=sqrt(100)*sqrt(100)=100.

2. Now let’s combine this with approximation theory, what is the sqrt(37)? Simply by noting that this is close to sqrt(36)=sqrt(9)*sqrt(4)=3*2=6 we have an approximate answer, the real answer is 6.08 by the way (see how close we can get!).

3. However, note that the above system is limited under certain contexts, first one must always be able to separate a number into multiples of 4s, 9s, 16s, 25s, 36s, and 49s. The result leaves one very limited with only two numbers of one decimal place to work with. But there are three special values that will greatly increase your range, those are the sqrt(2), sqrt(3) and the sqrt(10). Because any non-prime number can be produced by some combination of the multiples of 2 and 3, you can gain access to most numbers by the memorization of the basic even and odd multipliers sqrt(2) and the sqrt(3).

sqrt(2)=1.41~ 3/2
sqrt(3)=1.73~ 7/4
sqrt(10)=3.16~ 16/5

Now, if you have these simply have the sqrt(2) memorized you can access any even number that exists! Simply make use of method 1 multiple times and multiply all the 1.4s together. You can even combine 1 and 2 together in most circumstances by adding or subtracting 1 number to the system, but it is generally better to add than subtract to change the approximation to an even number because of the nature of this beast.
EX.

sqrt(17)~sqrt(16)=4, or ~ sqrt(18)=sqrt(9)*sqrt(2)=4.2

Actually

sqrt(17)=4.12

Which is slightly closer to 4.2 than it is to 4.0, though a half way point via a linear extrapolation of the points would really get you close!

When you add threes into the equation you can then gain access to any non-prime number that exists! I simply added 10 because of how useful it is when dealing with orders of magnitude. For instance, it is much easier to find sqrt(24000) by changing it to,

Sqrt(24000)=sqrt(24)*sqrt(100)*sqrt(10)=sqrt(3)*sqrt(2)*sqrt(2)*sqrt(2)*10*sqrt(10) then it is to find sqrt(240)*sqrt(100)=10*sqrt(2)*sqrt(2)*sqrt(2)*….sqrt(2). It just generally ends up paying to keep this number memorized, and for large orders of magnitude such as sqrt(24000) the nature of the beast will suffice with a simple sqrt(2)*sqrt(100)*sqrt(100)=141, the actual answer by the way is 155, (we’re not that far off!).

If you really wanted to get obsessed over sqrts, you could also memorize all the other numbers between 1 and 10. I’d suggest doing it to only two decimal places to avoid trouble, but mostly the above will get you through anything you’d need to do…

If you’re really nuts though, you can come up with equations to approximate sqrts. The Taylor series will work for certain things though if I remember correctly it only works between 0 and 1 for approximating square roots. Of course the above theories are still fine, so if one wanted to calculate the sqrt (32) one would only have to have an accurate taylor series that would yield the sqrt(0.32) Then by multiplying the answer by 10 (the sqrt(100)) you would have the sqrt(100)*sqrt(0.32)=sqrt(0.32)! I might note that I have a happy little book of equations I’ve etched out from hours of “playingâ€