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

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 $8 , \ldots . .$. Let's find the first digit with the same approach.

We can easy compute ${8.1}^{2}$ take you pencil and verify that $8.1 \cdot 8.1 = 65.61$. 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 $8.06225775 \ldots .$