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

1 Answer
George C.
Oct 18, 2015

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

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