# How do you find a one-decimal place approximation for #root3 15#?

##### 1 Answer

Oct 23, 2015

Find

#### Explanation:

Note that

So

To find an approximation for the cube root of a number

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

In our case

Then:

#a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 2.5 + (15 - 2.5^3)/(3*2.5^2)#

#=2.5 + (15-15.625)/18.75 = 2.5 - 0.625/18.75 = 2.4dot(6)dot(6)#

So our first approximation was already good to one decimal place.