# How do you find a one-decimal place approximation for #root3 30#?

##### 1 Answer

Oct 18, 2015

Use one step of Newton's method to find

#### Explanation:

To find the cube root of a number

#a_(i+1) = a_i + (n - a_i^3) / (3a_i^2)#

This is based on Newton's method for finding the zero of a function

#a_(i+1) = a_i - f(x)/(f'(x))#

In our case

Since

Then:

#a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 3+(30-27)/(3*3^2) = 3+3/27 = 3+1/9 = 3.dot(1)#

So to one decimal place