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

##### 1 Answer

Oct 18, 2015

Check

#### Explanation:

Halfway between

If you want more accuracy use something like a Newton Raphson type method:

To find approximations for

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

So in our case we might start with

#a_1 = (a_0^2+n)/(2a_0) = (5.5^2+30)/(2*5.5) = (30.25+30)/11 = 60.25/11 ~~ 5.477#