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

Dec 20, 2015

$3 \sqrt{- 8} \cdot i \sqrt{2} = - 12$

#### 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 $a , b \ge 0$:

If $a , b \ge 0$ then $\sqrt{a} \sqrt{b} = \sqrt{a b}$

Note carefully that $\sqrt{a} \sqrt{b} = \sqrt{a b}$ does not work if $a , b < 0$. For example:

$1 = \sqrt{1} = \sqrt{- 1 \cdot - 1} \ne \sqrt{- 1} \cdot \sqrt{- 1} = i \sqrt{1} \cdot i \sqrt{1} = {i}^{2} = - 1$

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

$3 \sqrt{- 8} \cdot i \sqrt{2}$

$= 3 i \sqrt{8} \cdot i \sqrt{2}$

$= 3 {i}^{2} \sqrt{8 \cdot 2}$

$= - 3 \sqrt{16}$

$= - 3 \cdot 4 = - 12$