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

1 Answer
Oct 14, 2015

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....