Question

How do you find the principal square root of 8?

Answer 1

For a non-negative real number, `n` (like `8`) the principal square root is the non-negative solution to `x^2=n`

The symbol `sqrtn` is used for the principal square root of `n`.

Thw pricipal square root of `8`, denoted `sqrt8` is the number whose sguare is `8`. There is no easy method for finding this number. We use successive approximation (or other iterative techniques) to get increasingly accurate approximations.

We do, however write `sqrt8` in "simpler" form `2sqrt2`

`sqrt8=sqrt(4*2)=sqrt4*sqrt2=2sqrt2`. Perhaps this is what you meant by "find"?

Answer 2

If, by "find" you mean get a decimal approximation, you could use: start with a number you know is close `3^2=9` and `2^2=4`, so we'll start with `3`

divide `8` by your last estimate: `8-:3=2.6667` (to 4 decimal places)
Average your previous estimate and the quotient:
`(3+2.6667)/2=2.8334` (rounding)

Repeat:
Divide: `8-:2.8334=2.8235`
Average: `(2.8334+2.8235)/2=2.8284`

Repeat
Divide: `8-:2.8284=2.8284`
Average `=2.8284`

`sqrt8~~2.8284` this estimate is accurate to 4 decimal places.
If you need more accuracy, start again and keep more places when rounding.