Question

What is the square root of 30?

Answer 1

You can't get an exact answer by hand, but you can estimate it using some tricks.

`sqrt(25) = 5`
`sqrt(36) = 6`

so it must be between `5` and `6`.

`30` is slightly less than halfway between `5` and `6`.

What you can begin with is assuming that because:

`30 - 25 = 5`
`36 - 25 = 11`
`5/11 = 45.45%`

... the square root of `30` is probably near `0.45 + 5 = 5.45`. What you are assuming then is:

`x + 0.45*[(x+1)-x] ~~ x^2 + 0.45*[(x+1)^2 - x^2]`

or more specifically,

`5 + 0.45*(6-5) ~~ 25 + 0.45*(36 - 25)`

Actually, it's not a bad bet. The actual square root is about `5.477`.

Answer 2

`30 = 2*3*5` has no square factors, so it is not possible to simplify `sqrt(30)`.

You can calculate an approximation by hand as shown below...

Explanation:

I explained my favourite method (a sort of Newton Raphson method) for approximating square roots of integers in an answer to the following question:

How do you find the square root 28?

Given an integer `n`, choose integers `p_0` and `q_0` so `p_0/q_0` is a reasonable first approximation to `sqrt(n)`.

Then iterate using the formulas:

`p_(i+1) = p_i^2 + n q_i^2`

`q_(i+1) = 2 p_i q_i`

If the resulting values of `p_(i+1)` and `q_(i+1)` have a common factor, then divide both by that factor before the next iteration.

The successive pairs `p_i`, `q_i` provide a sequence of rational approximations `p_i/q_i` to `sqrt(n)` that converge quite rapidly.

For our example, let `n = 30`, `p_0 = 11`, `q_0 = 2` (using an initial approximation of `5.5` since `30` is about halfway between `5^2 = 25` and `6^2 = 36`).

`p_1 = p_0^2 + n q_0^2 = 11^2 + 30*2^2 = 121 + 120 = 241`

`q_1 = 2 p_0 q_0 = 2 * 11 * 2 = 44`

This would give `sqrt(30) ~~ 241/44 = 5.47dot(7)dot(2)`

`p_2 = p_1^2 + n q_1^2 = 241^2 + 30*44^2 = 58081 + 58080 = 116161`

`q_2 = 2 p_1 q_1 = 2*241*44 = 21208`

This gives `sqrt(30) ~~ 116161/21208 ~~ 5.477225575`