# How do you find a one-decimal place approximation for #root5 26#?

##### 1 Answer

Use one step of Newton's method to find

#### Explanation:

To find an approximation for the

#a_(i+1) = a_i + (n - a_i^5)/(5 a_i^4)#

In our case let

Then:

So

If we want more accuracy, just apply the formula again to get

This method is a particular case of Newton's method for finding a zero of a continuous differentiable function

In our case