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

1 Answer
Oct 14, 2015

Answer:

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....#