Question

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

Answer 1

`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`