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

1 Answer
Oct 18, 2015

Answer:

#30# is about half way between #25 = 5^2# and #36 = 6^2#, so #5.5# might be a good approximation.

Check #5.5^2 = 30.25# is pretty close, so yes.

Explanation:

Halfway between #25=5^2# and #36=6^2# is #30.5# not #30#, but the parabola is also curved below the line joining #(5, 25)# and #(6, 36)#, so #5.5# is a better approximation for #sqrt(30)# than for #sqrt(31)#.

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

To find approximations for #sqrt(n)#, pick a starting approximation #a_0# and iterate using the formula:

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

So in our case we might start with #a_0 = 5.5#, then improve it by applying the formula:

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