Question

How do you find a one-decimal place approximation for `root2 65`?

Answer 1

As a one-decimal digit approximation, `sqrt(65)` is between `8` and `8.1`.

Explanation:

I am assuming of course that you can't use a calculator, otherwise the answer is: read it :)

So, you need an approximation, and a very simple way to find it is to try various numbers (of course with some wisdom, not just trying random numbers 'till you're lucky!): when you find a number which is smaller than the one you need, and one which is bigger, then surely the number you're looking for is between them.

Let's start searching the right integer: since `8^2=64` and `9^2=81`, `sqrt(65)` must be between `8` and `9`, i.e. a number of the form `8,.....`. Let's find the first digit with the same approach.

We can easy compute `8.1^2` take you pencil and verify that `8.1*8.1=65.61`. This is already larger than what we need! And since `8=8.0` is the lower approximation, we're done! `sqrt(65)` must be between `8` and `8.1`.

I hope that at this point the routine is clear for any number of digits you like.

Just for completion, the "actual" square root is `8.06225775....`