Question

How do you simplify `3sqrt(-8) • isqrt2`?

Answer 1

`3sqrt(-8)*isqrt(2) = -12`

Explanation:

Any non-zero number has two square roots, but when we write `sqrt(x)` we want it to indicate one of the roots unambiguously, which is called the principal square root.

If `x > 0` then both of its square roots are Real. By definition the principal square root `sqrt(x)` is the positive one.

If `x < 0` then both of its square roots are pure imaginary. By definition the principal square root `sqrt(x)` is the one with a positive coefficient of `i`, that is `i sqrt(-x)`.

If `x < 0` then `sqrt(x) = i sqrt(-x)`

With these definitions of `sqrt(x)`, the following identity holds when `a, b >= 0`:

If `a, b >= 0` then `sqrt(a)sqrt(b) = sqrt(ab)`

Note carefully that `sqrt(a)sqrt(b) = sqrt(ab)` does not work if `a, b < 0`. For example:

`1 = sqrt(1) = sqrt(-1 * -1) != sqrt(-1) * sqrt(-1) = i sqrt(1) * i sqrt(1) = i^2 = -1`

With this in mind, we exercise some care to write:

`3sqrt(-8)*isqrt(2)`

`=3isqrt(8)*isqrt(2)`

`=3i^2 sqrt(8*2)`

`= -3sqrt(16)`

`=-3*4=-12`