How do you simplify `sqrt42` rounded to the nearest tenths place?

If you ask a calculator to find `sqrt(42)` then it will give you an approximation like:

`sqrt(42) ~~ 6.4807406984`

So this is between `6.4` and `6.5`, but much closer to `6.5` than `6.4`.

So rounded to the nearest tenths place, `sqrt(42) ~~ 6.5`

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How would you find this without a calculator?

Note that:

`6*6 = 36`

`6*7 = 42`

`7*7 = 49`

`42` is nearly half way between `6^2 = 36` and `7^2 = 49`

So we expect `sqrt(42)` to be about halfway between `6` and `7`.

Looking at the graph of `y = x^2` for `x` between `6` and `7` and drawing a chord we see:

graph{(y-x^2)(y - (36+13(x-6))) = 0 [5.8, 7.2, 32, 51]}

By the time we get to `x in [6, 7]` the parabola is pretty straight, so we expect this linear interpolation to be good.

In fact we find:

`6.5^2 = 42.25`

Using Newton's method, the error in the approximation will be about:

`0.25 / (2*6.5) = 0.25/13 ~~ 0.02`

This is much smaller than `0.1` so will not affect the tenths place.

Another way of looking at this is continued fractions.

We find that any number of the form `n(n+1)` has a square root given by:

`sqrt(n(n+1)) = [n;bar(2,2n)] = n + 1/(2+1/(2n+1/(2+1/(2n+1/(2+...)))))`

In our example `42 = 6(6+1)`, so

`sqrt(42) = 6+1/(2+1/(12+1/(2+1/(12+1/(2+...)))))`

Truncating this, we can find rational approximations, such as:

`sqrt(42) ~~ [6;2] = 6+1/2 = 6.5`

`sqrt(42) ~~ [6;2,12] = 6+1/(2+1/12) = 6+12/25 = 6.48`