# Square roots

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how do i find the square root of a non-perfect square (sorry if its the wrong sub for this)

In: Mathematics

This site has a good explanation of the method I was taught by my math teacher who “didn’t believe in calculators!

https://byjus.com/maths/square-root-finder/

There are many methods. They tend to be recursive. Let me give you a simple one.

I want the square root of ten. I’m gonna use the ‘halfway guess and check’ method.

I know it’s bigger than zero and less than ten. Halfway between is five. Let’s square five. Twenty-five. Too big.

I know it’s bigger than 0 and less than 5. Halfway between is 2.5. Let’s square that. 6.25. Too small.

We know it’s bigger than 2.5 but less than 5. Halfway between is 3.75. Square that. 14.0625. Too big.

So it must be bigger than 2.5 but less than 3.75.

I’m done narrating. I’ll shorthand the steps.

3.125^2 = 9.76

Too small.

3.125<x<3.75

3.4375^2 = 11.816

Too big.

3.125<x<3.4375

3.28125^2 = 10.7666

Too big.

3.125<x<3.28125

I could keep doing this forever. I’ll get pretty close eventually. More efficient methods exist, but this one is nice and simple. Computers can do it ten thousand times in an instant, and that’s how my phone found the square root of ten (to within several decimal places): 3.16227766017

It can be just guess work.

If you’re doing it in your head then pick the two perfect squares either side. Like for sqrt 110 I’d say I know it’s between 10 and 11 (100 and 121 respectively). So a first guess would be 10.5. try that and second guess will.either be a bit higher or lower

There are more sophisticated methods to guess better. Ultimately they’re stil just approximations.

One other way is fancy maths simplification. So sqrt(200) = sqrt(100*2)=sqrt(100)*sqrt(2)=10*sqrt(2). So if I know sqrt(2) I can also tell you sqrt(200). It’s still guesswork to get the first one, but from there it gets easier.

I have a feeling there is some fancy work you can do with log tables, but for the life of me I can work out how to type it when I’m on mobile.

Edit: I think I have the log answer.

So, say x^2 = 200.
Log of both sides: log(X^2) = log(200)
Simplify: 2log(X)=log(200).

Log(X)= log(200)/2

Solve for X: X= 10^(log(200)/2)

Some clever people have calculated all the log values, it’s in a log table. So you can just read the number off for log(200). Then you have a clear sum to calculate.