How do you find a one-decimal place approximation for #sqrt 30#?
1 Answer
Oct 18, 2015
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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#